calki 10 id 107947 Nieznany

background image

Całki

Made by spajder

Ostatnia aktualizacja: 2006-07-08

Jeśli zauważysz błędy, napisz na adres :

Spajder1987@poczta.interia.pl

1.

Wprost z definicji:

( )

(

) ( )

+

=

+

=

+

=

+

=

+

+

=

+

=

+

=

+

=

+

=

+

+

=

+

=

+

=

+

=

+

=

+

=

+

=



+

+

=

+

=

+

=

+

=

+

=

+

,

1

1

,

;

arctgh

1

1

,

1

;

artgh

1

arcosh

1

arsinh

1

ctg

sinh

tgh

cosh

sinh

cosh

cosh

sinh

arctg

1

ctg

sin

tg

cos

sin

cos

arcsin

1

cos

sin

ln

1

;

1

1

;

|

|

ln

)

(

)

(

'

2

2

2

2

2

2

2

2

2

2

1

x

C

x

x

dx

x

C

x

x

dx

C

x

x

dx

C

x

x

dx

C

h

x

dx

C

x

x

dx

C

x

xdx

C

x

xdx

C

x

x

dx

C

x

x

dx

C

x

x

dx

C

x

xdx

C

e

dx

e

C

x

x

dx

C

x

xdx

C

a

a

dx

a

n

C

n

x

n

C

x

dx

x

C

ax

adx

C

x

dx

C

x

f

dx

x

f

x

x

x

x

n

n

background image

2.

Wyprowadzenia:

1)

C

a

x

a

C

t

a

t

dt

a

a

x

dx

a

a

a

dx

dt

a

x

t

a

x

dx

a

x

a

dx

+

=

+

=

+

=

+

=

=

=

=

+

=

+

arctan

1

arctan

1

1

1

)

(

1

1

1

)

(

1

1

2

2

2

2

2

2

2)

założenia :

U

Z

k

k

R

x

+

2

\

π

π

+

=

+

=

=

=

=

=

=

=

C

x

C

t

t

dt

dx

x

x

xdx

dt

x

t

dx

x

x

xdx

|

cos

|

ln

|

|

ln

cos

sin

sin

cos

cos

sin

tg

3)

założenia:

U

Z

k

k

R

x

π

\

+

=

+

=

=

=

=

=

=

C

x

C

t

t

dt

xdx

dt

x

t

dx

x

x

xdx

|

sin

|

ln

|

|

ln

cos

sin

sin

cos

ctg


4)

założenia :

+

R

x

+

=

+

=

=

=

=

=

=

=

=

C

x

x

C

x

x

x

dx

x

x

dx

x

x

x

x

x

v

x

du

dv

x

u

dx

x

)

1

(ln

ln

ln

1

ln

1

1

ln

ln

5)

założenia:

0

a

x



+

=

+

=



+

=

+

=

=

=

=

=

=

1

;

)

(

1

1

;

|

|

ln

1

;

1

1

;

|

|

ln

)

(

)

(

1

1

n

C

a

x

n

A

n

C

a

x

A

n

C

t

n

A

n

C

t

A

t

dt

A

dx

dt

a

x

t

a

x

dx

A

a

x

Adx

n

n

n

n

n


6)

założenia

0

;

0

+

c

a

b

ax

=

+

+

=

+

+

=

+

+

+

=

+

+

=

+

+

a

b

x

dx

ac

bc

ad

a

c

dx

a

c

dx

a

b

x

ac

bc

ad

a

c

dx

a

b

x

a

b

c

d

a

b

x

a

c

dx

a

b

x

c

d

x

a

c

dx

b

ax

d

cx

)

1

(

C

a

b

x

a

bc

ad

x

a

c

C

t

a

bc

ad

x

a

c

t

dt

a

bc

ad

dx

a

c

dx

dt

a

b

x

t

+

+

+

=

+

+

=

+

=

=

+

=

=

|

|

ln

|

|

ln

2

2

2

7)

założenia:

)

,

3

(

Z

n

=

=

=

=

=

=

=

x

v

x

x

n

du

x

dv

x

u

xdx

x

xdx

n

n

n

n

cos

cos

sin

)

1

(

sin

sin

sin

sin

sin

2

1

1

=

+

=

+

xdx

x

n

x

x

xdx

x

n

x

x

n

n

n

n

2

2

1

2

2

1

cos

sin

)

1

(

sin

cos

cos

sin

)

1

(

cos

sin

=

+

xdx

x

x

n

x

x

n

n

)

sin

1

(

sin

)

1

(

sin

cos

2

2

1

background image

=

+

xdx

dx

x

n

xdx

n

x

x

n

n

n

n

sin

sin

)

1

(

sin

)

1

(

sin

cos

2

1

stąd:

+

=

dx

x

n

n

x

x

n

xdx

n

n

n

2

1

sin

1

sin

cos

1

sin

8)

założenia

)

,

3

(

Z

n

=

=

=

=

=

=

=

x

v

x

x

du

x

dv

x

u

xdx

x

xdx

n

n

n

n

sin

cos

sin

cos

cos

cos

cos

cos

2

1

1

=

+

=

+

dx

x

x

n

x

x

xdx

x

n

x

x

n

n

n

n

)

cos

1

(

cos

)

1

(

cos

sin

sin

cos

)

1

(

cos

sin

2

2

1

2

2

1

+

+

xdx

n

xdx

n

x

x

n

n

n

cos

)

1

(

cos

)

1

(

cos

sin

2

1

stąd:

+

=

xdx

n

n

x

x

n

xdx

n

n

n

2

1

cos

1

cos

sin

1

cos

9)

założenia:

0

2

2

>

x

a

+

=

+

=

=

=

=

=

=

=

C

a

x

C

t

t

dt

a

x

dx

a

a

dx

dt

a

x

t

a

x

dx

a

x

a

dx

arcsin

arcsin

1

)

(

1

1

)

(

1

1

2

2

2

2

2


10)

założenia:

0

2

2

x

a

=

=

=

2

2

2

2

2

2

2

2

2

2

2

x

a

xdx

x

x

a

dx

a

dx

x

a

x

a

dx

x

a

+

=

=

=

=

=

dx

x

a

x

a

x

x

a

dx

a

x

a

v

du

x

a

x

dv

x

u

2

2

2

2

2

2

2

2

2

2

2

1

stąd:

C

x

a

x

a

x

a

x

a

x

x

a

dx

a

dx

x

a

+

+

=

=

+

=

2

2

2

2

2

2

2

2

2

2

2

arcsin

2

9

2

2

11)

założenia :

0

>

k

+

+

+

=

+

=

=

=

+

+

+

+

=

+

+

=

+

+

=

=

+

C

k

x

x

C

t

t

dt

t

dt

k

x

dx

dx

k

x

k

x

x

dx

k

x

x

dt

k

x

x

t

k

x

dx

|

|

ln

|

|

ln

)

1

(

2

2

2

2

2

2

2

12)

założenia :

0

>

k

=

+

=

=

+

=

=

=

+

+

+

=

+

+

=

+

k

x

v

du

k

x

x

dv

x

u

k

x

dx

k

dx

k

x

x

x

dx

k

x

k

x

dx

k

x

2

2

2

2

2

2

2

1

dx

k

x

dx

k

x

k

x

x

k

x

dx

k

+

=

+

+

+

+

=

2

2

2

2

background image

stąd:

C

k

x

x

k

x

x

k

k

x

x

k

x

dx

k

dx

k

x

+

+

+

+

+

=

=

+

+

+

=

+

2

2

2

2

2

2

|

|

ln

2

11

2

2

13)

założenia:

0

)

(

x

f

+

=

+

=

=

=

=

=

C

x

f

C

t

t

dt

dx

x

f

dt

x

f

t

dx

x

f

x

f

|

)

(

|

ln

|

|

ln

)

(

'

)

(

)

(

)

(

'

14)

założenia:

n

m

±

=

+

=

+

=

=

dx

n

m

x

dx

n

m

x

y

x

y

x

y

x

nxdx

mx

)]

(

cos[

2

1

)]

(

cos[

2

1

)

cos(

)

[cos(

2

1

sin

sin

sin

sin

=

+

=

+

=

=

=

du

n

m

du

n

m

x

u

dx

n

m

dt

n

m

x

t

)

(

)

(

)

(

)

(

=

+

+

+

dx

n

m

n

m

x

n

m

dx

n

m

n

m

x

n

m

)

)](

(

cos[

)

(

2

1

)

)](

(

cos[

)

(

2

1

C

n

m

n

m

x

n

m

n

m

x

C

n

m

u

n

m

t

udu

n

m

tdt

n

m

+

+

+

=

+

+

=

+

)

(

2

)]

(

sin[

)

(

2

)]

(

sin[

)

(

2

sin

)

(

2

sin

cos

)

(

2

1

cos

)

(

2

1

15)

założenia:

n

m

±

=

+

+

=

=

)

cos(

)

[cos(

2

1

cos

cos

cos

cos

y

x

y

x

y

x

nxdx

mx

=

=

=

+

=

+

=

=

+

+

dx

n

m

du

n

m

x

u

dx

n

m

dt

n

m

x

t

dx

n

m

x

dx

n

m

x

)

(

)

(

)

(

)

(

)]

(

cos[

2

1

)]

(

cos[

2

1

=

+

+

+

+

dx

n

m

n

m

x

n

m

dx

n

m

n

m

x

n

m

)

)](

(

cos[

)

(

2

1

)

)](

(

cos[

)

(

2

1

C

n

m

n

m

x

n

m

n

m

x

C

n

m

u

n

m

t

udu

n

m

tdt

n

m

+

+

+

+

=

+

+

+

=

+

+

)

(

2

)]

(

sin[

)

(

2

)]

(

sin[

)

(

2

sin

)

(

2

sin

cos

)

(

2

1

cos

)

(

2

1



16)

założenia:

n

m

±

=

+

+

=

+

+

=

=

dx

n

m

x

dx

n

m

x

y

x

y

x

x

x

nxdx

mx

)]

(

sin[

2

1

)]

(

sin[

2

1

)

sin(

)

[sin(

2

1

cos

sin

cos

sin

=

=

=

+

=

+

=

dx

n

m

du

n

m

x

u

dx

n

m

dt

n

m

x

t

)

(

)

(

)

(

)

(

=

+

+

+

+

dx

n

m

n

m

x

n

m

dx

n

m

n

m

x

n

m

)

)](

(

sin[

)

(

2

1

)

)](

(

sin[

)

(

2

1

=

+

=

+

+

)

(

2

cos

sin

)

(

2

1

sin

)

(

2

1

n

m

t

udu

n

m

tdt

n

m

C

n

m

n

m

x

n

m

n

m

x

C

n

m

u

+

+

+

+

=

+

)

(

2

)]

(

cos[

)

(

2

)]

(

cos[

)

(

2

cos

17)

założenia :

0

m

background image

+

=

+

=

=

=

=

=

=

C

m

mx

C

m

t

tdt

m

dx

m

x

m

mdx

dt

mx

t

mxdx

sin

sin

cos

1

cos

1

cos

18)

założenia :

0

m

C

m

mx

C

m

t

tdt

m

mdx

x

m

mdx

dt

mx

t

mxdx

+

=

+

=

=

=

=

=

=

cos

cos

sin

1

sin

1

sin

19)

założenia :

Z

k

k

mx

m

+

;

2

;

0

π

π

=

+

=

=

=

=

=

=

=

C

m

t

t

dt

m

dx

mx

mx

m

m

mx

m

dt

mx

t

dx

mx

mx

mxdx

|

|

ln

1

cos

sin

1

sin

cos

cos

sin

tg

C

m

mx

+

|

cos

|

ln

20)

założenia :

Z

k

k

mx

m

;

;

0

π

C

m

t

t

dt

m

dx

mx

mx

m

m

mxdx

m

dt

mx

t

dx

mx

mx

mxdx

+

=

=

=

=

=

=

=

|

|

ln

1

sin

cos

1

cos

sin

sin

cos

ctg

C

m

mx

+

|

sin

|

ln

21)

=




+

+

=

+

=

=

+

=

=

+

=

=

+

+

1

;

)

1

(

1

;

ln

1

)

(

1

)

(

1

n

C

n

a

t

n

C

a

t

dt

t

a

adx

b

ax

a

adx

dt

b

ax

t

dx

b

ax

n

n

n

n




+

+

+

=

+

+

+

1

;

)

1

(

)

(

1

;

)

ln(

1

n

C

n

a

b

ax

n

C

a

b

ax

n

22)

założenia:

0

cos

;

0

+

x

b

a

b

+

+

=

+

=

=

+

=

=

+

=

=

+

C

b

x

b

a

C

t

b

t

dt

b

x

b

a

xdx

b

b

x

b

dt

x

b

a

t

dx

x

b

a

x

|

cos

|

ln

|

|

ln

1

1

cos

cos

1

sin

cos

cos

sin

23)

założenia:

+

R

b

a,

=

+

=

+

=

=

+

=

=

+

=

=

+

C

t

b

C

t

b

dt

t

b

bdx

bx

a

b

bdx

dt

bx

a

t

dx

bx

a

2

3

2

3

3

2

2

3

1

1

1

C

bx

a

b

+

+

3

)

(

3

2


24)

założenia:

0

4

;

0

2

<

+

+

q

p

q

px

x



+

+

+

=

+

+

+

=



+

=

+

=

=

+

=

+

+

=

=

+

+

+

1

;

1

)

(

1

;

|

|

ln

1

;

1

1

;

|

|

ln

)

2

(

)

(

2

1

2

2

1

2

2

n

C

n

q

px

x

n

C

q

px

x

n

C

n

t

n

C

t

t

dt

dx

p

x

dt

q

px

x

t

dx

q

px

x

p

x

n

n

n

n

background image

25)

+

+

=

+

+

=




+

=

+

=

=

+

=

=

+

=

=

+

1

;

2

2

)

1

(

1

;

|

1

|

ln

2

1

1

;

)

1

(

2

1

;

|

|

ln

2

1

2

1

)

1

(

2

2

1

2

1

)

1

(

1

2

2

1

2

2

2

n

C

n

x

n

C

x

n

C

n

t

n

C

t

t

dt

x

xdx

xdx

dt

x

t

x

xdx

n

n

n

n

n

26)

założenia:

1

;

>

n

Z

n

+

+

+

=

+

+

=

+

n

n

n

n

x

dx

x

x

x

dx

x

x

x

x

dx

)

1

(

)

1

(

1

)

1

(

1

)

1

(

2

2

2

2

2

2

2

2

Oznaczmy

+

=

n

n

x

dx

I

)

1

(

2

, więc

+

=

n

n

n

x

dx

x

I

I

)

1

(

1

.

26

2

2

1

=

+

=

=

+

=

=

=

+

=

+

1

2

2

2

2

2

)

1

)(

2

2

(

1

1

)

1

(

)

1

(

)

1

(

n

n

n

n

x

n

v

du

x

x

dv

x

u

x

xdx

x

x

dx

x

=

+

+

+

=

+

+

+

1

2

1

2

1

2

1

2

)

1

(

)

2

2

(

1

)

1

)(

2

2

(

)

1

)(

2

2

(

)

1

)(

2

2

(

n

n

n

n

x

dx

n

x

n

x

x

n

dx

x

n

x

1

1

2

)

2

2

(

1

)

1

(

2

2

1

+

+

=

n

n

I

n

x

x

n

podstawiając do <26.1>:

1

1

2

1

2

2

1

)

1

(

2

2

1

+

+

=

n

n

n

n

I

n

x

x

n

I

I

a ponieważ:

+

=

n

n

x

dx

I

)

1

(

2

to:

+

+

+

=

+

1

2

1

2

2

)

1

(

2

2

3

2

)

1

(

2

2

1

)

1

(

n

n

n

x

dx

n

n

x

x

n

x

dx

27)

założenie:

+

<

=

Z

n

q

p

;

0

4

2

=

+

=

+

=

+

+

+

=

+

+

n

n

n

n

p

x

dx

q

p

p

x

dx

p

q

p

px

x

dx

q

px

x

dx

]

4

)

2

[(

]

4

4

)

2

[(

)

4

4

(

)

(

2

2

2

2

2

2

2

=

+

Λ

=

=

=

=

+

n

n

n

n

p

x

dx

dx

dt

p

x

t

p

x

dx

]

1

)

2

[(

2

2

)

4

(

2

2

]

1

)

2

[(

)

4

(

2

2

+

n

n

t

dt

)

1

(

)

4

(

2

2

1

tutaj należy skorzystać ze wzoru <26> (dla n>1).

28)

Założenia:

0

4

;

2

<

+

q

p

Z

n

background image

=

+

+

+

+

+

+

=

+

+

+

+

=

+

+

+

dx

q

px

x

p

A

B

A

dx

q

px

x

p

x

A

dx

q

px

x

p

p

A

B

x

A

dx

q

px

x

B

Ax

n

n

n

n

)

(

2

2

)

(

2

2

)

(

2

2

2

)

(

2

2

2

2

=

+

+

+

=

+

=

+

+

=

=

n

n

q

px

x

dx

A

Ap

B

A

t

dt

A

dx

p

x

dt

q

px

x

t

)

(

2

2

2

)

2

(

2

2

=

=

=

=

=




+

+

=

+

+

+

2

2

2

1

2

4

2

4

2

27

1

;

)

(

)

2

(

)

1

(

2

1

;

)

(

)

2

(

|

|

ln

2

q

p

dx

du

p

q

p

x

u

n

q

px

x

dx

Ap

B

n

At

n

q

px

x

dx

Ap

B

t

A

n

n

n

n

(

)

+

+

+

+

+

+

+

+

=

n

n

n

u

du

Ap

B

n

q

px

x

A

C

p

q

p

x

p

q

Ap

B

q

px

x

A

1

)

4

)(

2

(

)

1

(

2

)

(

4

2

arctg

)

2

4

)(

2

(

)

ln(

2

2

2

1

1

2

2

2

2

tutaj zastosować wzór <26>

29)

Założenia:

+

R

b

C

b

k

x

arctg

b

C

b

t

arctg

b

b

t

dt

dx

dt

k

x

t

b

k

x

dx

+

=

+

=

=

+

=

=

=

=

+

1

1

1

)

(

2

2

30)

Założenia :

0

sin

x

+

=

+

=

=

=

=

=

=

=

C

x

C

t

t

dt

x

dx

dt

x

tg

t

x

x

dx

x

x

dx

x

dx

|

2

tg

|

ln

|

|

ln

2

cos

2

2

cos

2

tg

2

2

cos

2

sin

2

sin

2

2

31)

Założenia :

0

cos

x

=

+

=

=

=

=

+

=

=

+

=

+

=

=

C

t

t

dt

dx

dt

x

t

x

dx

x

x

x

dx

|

2

tg

|

ln

30

sin

2

)

2

sin(

)

2

sin(

cos

cos

π

π

π

C

x

+

+

|

)

2

4

tg(

|

ln

π

32)

Założenia:

0

cos

sin

x

x

C

x

C

t

t

dt

dx

dt

x

t

x

dx

x

dx

x

x

dx

+

=

+

=

=

=

=

=

=

=

=

|

tg

|

ln

|

2

tg

|

ln

30

sin

2

2

2

sin

2

2

sin

2

1

cos

sin

33)

Założenia :

0

cos

sin

x

x

=

=

=

+

=

x

dx

x

dx

x

x

xdx

x

x

xdx

x

x

dx

x

x

x

x

dx

2

2

2

2

2

2

2

2

2

2

2

2

2

2

sin

cos

cos

sin

cos

cos

sin

sin

cos

sin

)

cos

(sin

cos

sin

C

x

x

+

ctg

tg

background image

34)

Założenia :

U

Z

k

k

R

x

+

π

π

2

\

+

=

=

=

=

C

x

x

x

xdx

x

dx

x

dx

x

x

xdx

xdx

tg

cos

cos

cos

cos

)

cos

1

(

cos

sin

tg

2

2

2

2

2

2

2

2

35)

Założenia :

U

Z

k

k

R

x

π

\

+

=

=

=

=

C

x

x

x

xdx

x

dx

x

dx

x

x

xdx

xdx

ctg

sin

sin

sin

sin

)

sin

1

(

sin

cos

ctg

2

2

2

2

2

2

2

2

36)

założenia :

U

Z

k

k

R

x

n

Z

n

+

>

π

π

2

\

;

2

;

=

=

=

=

=

dx

x

x

x

x

x

dx

x

x

x

dx

x

tg

x

dx

x

n

n

n

n

n

)

1

cos

1

(

tg

cos

cos

1

tg

cos

sin

tg

tg

tg

2

2

2

2

2

2

2

2

2

=

=

=

=

=

=

xdx

t

n

xdx

dt

t

x

dx

dt

x

t

xdx

x

dx

x

n

n

n

n

n

n

2

1

2

2

2

2

2

2

tg

1

1

tg

cos

tg

tg

cos

tg

xdx

n

n

n

2

1

tg

tg

1

1

37)

założenia:

U

Z

k

k

R

x

n

Z

n


>

π

\

;

2

;

=

=

=

=

dx

x

x

x

dx

x

x

x

dx

x

x

xdx

n

n

n

n

2

2

2

2

2

2

2

2

sin

sin

1

ctg

sin

cos

ctg

ctg

ctg

ctg

=

=

=

=

=

x

dx

dt

x

t

dx

x

x

dx

x

dx

x

x

n

n

n

2

2

2

2

2

2

sin

ctg

ctg

sin

ctg

)

1

sin

1

(

ctg

=

=

xdx

x

n

xdx

t

n

xdx

dt

t

n

n

n

n

n

n

2

1

2

1

2

2

ctg

ctg

1

1

ctg

1

1

ctg

38)

C

c

x

x

c

cx

cxdx

c

c

cx

x

cx

c

v

du

cx

dv

x

u

cxdx

x

+

=

+

=

=

=

=

=

=

cos

sin

cos

1

cos

cos

1

1

sin

sin

2

39)

założenia:

0

sin

cx

C

cx

c

C

t

c

t

dt

c

cdx

dt

cx

t

cx

dx

+

=

+

=

=

=

=

=

=

|

2

tg

|

ln

1

|

2

tg

|

ln

1

30

sin

1

sin

40)

założenia:

{

}

1

,

2

n

background image

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

(

) (

)

(

)

(

)(

)

(

)

(

)

(

)(

)

(

)

(

)(

)(

)

C

b

ax

n

n

a

b

x

n

a

C

n

n

a

b

ax

x

n

a

b

ax

C

n

n

a

b

ax

b

ax

x

n

a

C

n

n

a

b

ax

n

a

b

ax

x

C

n

t

n

a

n

a

b

ax

x

dt

t

n

a

n

a

b

ax

x

adx

dt

b

ax

t

n

a

dx

b

ax

n

a

b

ax

x

n

a

b

ax

v

du

b

ax

dv

x

u

dx

b

ax

x

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

+

+

+

+

=

+

+

+

+

+

=

+

+

+

+

+

+

=

+

+

+

+

+

+

=

+

+

+

+

+

+

=

+

+

+

=

=

+

=

=

+

+

+

+

=

+

+

=

=

+

=

=

=

+

+

+

+

+

+

+

+

+

+

+

+

+

+

1

2

2

1

2

2

1

2

2

1

2

2

1

1

2

1

1

1

1

2

1

1

2

1

2

2

1

2

2

1

1

2

1

1

1

1

1

1

1

)

1

(

)

1

(

1

41)

założenie :

0

+

b

ax

(

)

C

b

ax

a

b

a

b

a

x

C

b

ax

a

b

a

b

ax

C

a

a

b

a

t

t

dt

a

b

dt

a

at

dt

b

t

a

a

b

t

x

adx

dt

b

ax

t

b

ax

xdx

+

+

+

=

+

+

+

=

+

=

=

=

=

=

+

=

=

+

ln

ln

ln

1

1

2

2

2

2

2

2

2

2

Ponieważ

2

a

b

jest stałą można ją włączyć do stalej całkowania. Tak więc:

C

b

ab

b

a

a

x

b

ax

xdx

+

+

=

+

ln

2

42)

założenia:

0

+

b

ax

(

)

(

)

(

)

C

b

ax

a

b

ax

a

t

a

t

a

C

t

a

t

a

t

dt

a

t

dt

a

at

dt

b

t

a

a

b

t

x

adx

dt

b

ax

t

b

ax

xdx

+

+

+

+

=

+

=

+

=

=

=

=

=

+

=

=

+

ln

1

1

1

ln

1

1

1

ln

1

1

1

1

2

2

2

2

2

2

2

2

2

2

2

2

43)

założenia:

{ }

2

,

1

;

0

+

n

b

ax

(

)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)(

)(

)

(

)(

)(

)

(

)

(

)(

)(

)

(

)

(

)(

)(

)

C

b

ax

n

n

a

b

x

n

a

C

b

ax

n

n

a

b

x

n

a

C

b

ax

n

n

a

bn

b

bn

b

nax

ax

C

b

ax

n

n

a

bn

b

n

b

ax

C

n

n

a

n

b

n

t

t

C

n

t

a

b

n

t

a

t

a

b

dt

t

a

at

dt

b

t

a

a

b

t

x

adx

dt

b

ax

t

b

ax

xdx

n

n

n

n

n

n

n

n

n

t

n

+

+

=

+

+

=

+

+

+

+

=

+

+

+

+

=

+

=

+

=

=

=

=

=

+

=

=

+

1

2

1

2

1

2

1

2

2

1

1

2

2

2

2

1

2

2

1

1

1

2

1

1

2

2

1

2

2

1

1

2

2

1

1

2

1

1

1

44)

założenia:

0

+

b

ax

background image

( )

(

)

(

)

C

b

ax

b

b

ax

b

b

ax

a

C

t

b

bt

t

a

t

dt

b

dt

b

tdt

a

dt

t

b

tb

t

a

t

a

dt

b

t

a

a

b

t

x

adx

dt

b

ax

t

b

ax

dx

x

+

+

+

+

+

=

+

+

=





+

=

+

=

=

=

=

+

=

=

+

ln

2

2

1

ln

2

2

1

2

1

2

1

1

2

2

3

2

2

3

2

3

2

2

3

2

2

2

45)

założenia:

0

+

b

ax

(

)

(

)

C

b

ax

b

b

ax

b

b

ax

a

C

t

b

t

b

t

a

t

dt

b

t

dt

b

dt

a

dt

t

b

tb

t

a

t

a

dt

b

t

a

a

b

t

x

adx

dt

b

ax

t

b

ax

dx

x

+

+

+

+

=

+

=





+

=

+

=

=

=

=

+

=

=

+

2

3

2

3

2

2

3

2

2

2

3

2

2

2

2

2

ln

2

1

ln

2

1

2

1

2

1

1

46)

założenia:

0

+

b

ax

(

)

(

)

(

)

C

b

ax

b

b

ax

b

b

ax

a

C

t

b

t

b

t

a

t

dt

b

t

dt

b

t

dt

a

dt

t

b

tb

t

a

t

a

dt

b

t

a

a

b

t

a

adx

dt

b

ax

t

b

ax

dx

x

+

+

+

+

+

=

+

+

=





+

=

+

=

=

=

=

+

=

=

+

2

2

3

2

2

3

3

2

2

3

3

2

2

3

3

2

2

3

2

2

2

ln

1

2

2

ln

1

2

1

2

1

1

47)

założenia:

{

}

3

,

2

,

1

;

0

+

n

b

ax

(

)

(

)

[

]

(

)(

)

(

)(

)

(

)(

)

C

b

ax

n

b

b

ax

n

b

b

ax

n

a

C

n

t

b

n

bt

n

t

a

dt

t

b

dt

t

b

dt

t

a

t

b

bt

t

a

t

a

dt

b

t

a

a

b

t

x

adx

dt

b

ax

t

b

ax

dx

x

n

n

n

n

n

n

n

n

n

n

n

n

+

+

+

+

+

=

+

+

=

+

=

+

=

=

=

=

+

=

=

+

1

2

2

3

3

1

2

2

3

3

2

1

2

3

2

2

3

2

2

2

1

2

2

3

1

1

1

2

2

3

1

2

1

2

1

1

48)

założenia:

(

)

0

;

0

+

b

b

ax

x

(

)

+

b

ax

x

dx

Funkcję podcałkową rozbijam na sumę ułamków prostych:

background image

(

)

(

)

(

)

C

x

b

ax

C

b

ax

b

x

b

b

ax

dx

b

a

x

dx

b

b

ax

x

dx

b

a

B

a

b

B

a

b

x

b

A

Ab

x

Bx

b

ax

A

b

ax

B

x

A

b

ax

x

+

+

=

+

+

>=

=<

+

=

+

=

=

=

=

+

+

+

+

+

ln

ln

1

ln

1

21

1

:

Tak więa

1

Postawiam

1

1

:

0

Podstawiam

1

1


49)

założenia:

(

)

0

+

b

ax

abx

(

)

:

prostych

ulamków

sume

na

podcalkowa

funkcję

Rozbijam

2

+

b

ax

x

dx

(

)

(

) (

)

(

) (

)

1

1

1

1

1

1

:

Rozpisuj ę

1

podstawiam

1

1

:

0

podstawiam

1

1

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

+

+

+





+

+

+

+

+

+

+

+

+

=

=

=

=

+

+

+

+

+

+

+

+

x

b

a

Ab

x

b

a

Aa

x

b

a

x

b

a

Abx

Aax

x

b

a

b

ax

b

b

ax

Ax

b

a

C

a

b

C

a

b

x

b

B

Bb

x

Cx

b

ax

B

b

ax

Ax

b

ax

C

x

B

x

A

b

ax

x

stąd:

background image

(

)

C

x

b

ax

b

a

bx

C

b

ax

b

a

bx

x

b

a

b

ax

dx

b

a

x

dx

b

x

dx

b

a

b

ax

x

dx

b

a

A

b

a

Aa

+

+

+

=

+

+

+

>=

=<

+

+

+

=

+

=

+

=

ln

1

ln

1

ln

21

1

:

Tak więa

0

2

2

2

2

2

2

2

2

2

2

2

50)

założenia:

(

)

0

+

b

ax

abx

(

)

+

2

2

b

ax

x

dx

Funkcję podcałkową rozbijam na sumę ułamków prostych.

(

)

(

)

(

)

(

)

(

)

2

2

2

2

2

2

2

2

2

2

1

1

0

podstawiam

1

1

b

B

Bb

x

Dx

b

ax

Cx

b

ax

B

b

ax

Ax

b

ax

D

b

ax

C

x

B

x

A

b

ax

x

=

=

+

+

+

+

+

+

+

+

+

+

+

+

2

1

podstawiam

=

a

b

D

a

b

x

(

)

(

)

(

)

(

)

(

)

(

)

(

)

:

więw

a

1

2

2

2

1

1

2

2

1

2

1

2

1

1

1

:

podstawiam

2

2

2

2

3

2

2

2

2

2

3

2

2

2

2

2

3

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

+

+

+





+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

=

x

b

a

Ab

x

Cb

b

a

Aab

x

Ca

Aa

x

b

a

Cbx

Cax

x

b

a

x

b

a

x

Ab

Aabx

x

Aa

x

b

a

b

ax

Cx

b

abx

x

a

b

b

abx

x

a

Ax

x

b

a

b

ax

Cx

b

ax

b

b

ax

Ax

b

a

D

background image

3

2

3

3

3

2

2

2

2

2

2

2

:

pierwszego

rownania

do

podstawiam

2

2

:

wyliczam

równania

ostatniego

z

0

0

2

2

2

0

b

a

C

b

a

Ca

b

a

A

b

a

Ab

b

a

Ab

Cb

b

a

Aab

Ca

Aa

=

=

=

=



=

=

+

+

+

=

+

tak więc:

(

)

(

)

(

)

(

)

C

x

b

ax

b

x

ab

b

ax

b

a

C

b

ax

b

a

b

ax

b

a

x

b

x

b

a

b

ax

dx

b

a

b

ax

dx

b

a

x

dx

b

x

dx

b

a

b

ax

x

dx

+





+

+

+

=

+

+

+

+

+

>=

=<

+

+

+

+

+

=

+

ln

2

1

1

ln

2

1

ln

2

21

2

1

2

3

2

2

2

3

2

3

2

2

2

3

2

2

2

3

2

2

51)

założenia:

0

2

2

a

x

(

)(

)

+

=

a

x

a

x

dx

a

x

dx

2

2


rozkładam funkcję podcałkową na sumę ułamków prostych:

(

)(

)

(

) (

)

a

B

Ba

a

x

a

A

Aa

a

x

a

x

B

a

x

A

a

x

B

a

x

A

a

x

a

x

2

1

2

1

podstawiam

2

1

2

1

podstawiam

1

1

=

=

=

=

+

+

+

+

+

tak więc:

(

)

+

+

=

+

+

>=

=<

+

=

C

a

x

a

x

a

C

a

x

a

x

a

a

x

dx

a

a

x

dx

a

a

x

dx

ln

2

1

ln

ln

2

1

21

2

1

2

1

2

2

background image

52)

założenia :

0

2

2

a

x

( )

(

) ( )

(

)

(

) ( )

+

+

=

+

+

=

=

=

=

=

=

,

1

1

,

;

artgh

1

,

;

artgh

1

,

1

1

,

;

artgh

1

1

,

1

;

artgh

1

1

1

1

1

2

2

2

2

2

x

C

a

x

a

a

a

x

C

a

x

a

t

C

t

a

t

C

t

a

t

dt

a

a

dx

dt

a

x

t

a

x

dx

a

a

x

dx


53)

założenia:

0

2

2

x

a

a

(

)

C

x

a

C

t

dt

t

xdx

dt

x

a

t

dx

x

a

x

+

=

+

=

=

=

=

=

3

2

2

2

3

2

1

2

2

2

2

3

1

3

1

2

1

2

54)

+

+

>=

=<

+

=

+

=

C

x

x

dx

dx

x

dx

x

xdx

2

1

2

sin

4

1

17

2

1

2

cos

2

1

2

1

2

cos

cos

2

55)

C

x

x

dx

dx

x

dx

x

dx

x

+

>

=<

=

=

2

1

2

sin

17

2

1

2

cos

2

1

2

1

2

cos

sin

4

1

2

56)

założenia :

0

2

2

>

x

a

C

a

x

C

t

dt

x

a

dx

dt

a

x

t

t

a

x

x

a

dx

+

=

+

=

=

=

=

=

=

arcsin

arcsin

sin

2

2

2

2

57)

założenia :

0

2

2

>

x

a

(

)

(

)

C

a

x

a

x

a

x

a

C

a

x

a

x

a

x

a

C

x

a

x

a

C

u

t

a

du

u

dt

a

du

du

t

u

tdt

dt

dt

a

dt

t

dt

a

tdt

dt

a

dt

t

a

a

x

a

dx

dt

a

x

t

t

a

x

dx

x

a

x

a

dx

x

a

+

+

=

+





+

=

+





+

=

+

+

=

+

=

=

=

=

+

=

=

=

=

=

=

=

=

=

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1

2

arcsin

2

1

arcsin

cos

arcsin

sin

2

1

arcsin

2

1

2

arcsin

2

sin

4

1

arcsin

2

1

sin

4

1

2

1

cos

4

1

2

1

2

2

2

cos

2

1

2

1

2

2

cos

1

sin

sin

arcsin

sin

background image

C

x

a

x

a

x

a

C

a

x

ax

a

x

a

+

+

=

+

+

2

2

2

2

2

2

arcsin

2

1

2

arcsin

2

58)

(

)

( )

(

)

(

)

C

a

x

a

x

a

x

a

C

a

x

a

x

a

a

x

a

x

x

x

x

C

a

x

a

x

a

a

x

a

C

u

a

t

a

du

u

a

dt

a

dt

du

t

u

dt

a

dt

t

a

dt

a

x

x

tdt

a

dt

a

dt

t

a

a

a

x

dx

dt

a

x

t

t

a

x

dx

a

x

a

x

dx

a

x

+

+

+

=

+

+

+

=

+

=

=

=

+

+

=

+

+

=

+

=

=

=

+

=

=

=

+

=

+

=

+

=

=

=

=

+

+

=

+

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

arsinh

2

1

2

arsinh

2

1

arsinh

cosh

arsinh

sinh

arsinh

sinh

)

h

cosh(arsin

2

arsinh

2

sinh

4

2

cosh

4

2

2

2

2

2

cosh

2

2

1

2

cosh

sinh

sinh

sinh

arsinh

sinh

59)

C

a

x

C

t

dt

x

a

dx

dt

a

x

t

t

a

x

x

a

dx

+

=

+

=

=

+

=

=

=

=

+

arsinh

arsinh

sinh

2

2

2

2

60)

założenia:

0

;

0

2

2

>

x

x

a

(

)

(

)

=

+

+

=

+

+

=

+

+

>=

=

=

=<

+

+

=

+

+

>=

=

=<

+

+

>=

<

=

=

=

=

=

=

=

C

x

a

x

a

a

x

a

C

x

a

x

x

a

a

a

C

a

x

a

a

x

a

x

a

x

x

x

x

C

a

x

a

a

x

a

x

a

C

t

a

t

t

a

x

x

x

tg

C

t

a

t

a

tdt

a

t

dt

a

dt

t

a

t

a

a

x

a

dx

dt

a

x

t

t

a

x

dx

x

a

x

x

a

x

dx

x

a

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

ln

ln

1

1

1

ln

1

arcsin

cos

arcsin

sin

arcsin

cos

arcsin

sin

arcsin

cos

1

ln

cos

sin

cos

1

ln

sin

cos

1

2

cos

2

tg

ln

30

sin

sin

sin

sin

arcsin

sin


background image

(

)

(

)(

)

(

)

(

)

C

x

x

a

a

a

x

a

C

x

x

a

a

x

a

x

a

C

x

a

a

x

a

a

x

a

x

a

C

x

a

a

x

a

a

x

a

a

x

a

x

a

+

+

=

+

+

=

+

+

+

=

+

+

+

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

ln

ln

ln

ln

61)

0

;

0

2

2

>

x

x

a

C

a

x

a

a

x

a

a

x

a

C

a

t

a

t

a

t

C

a

u

a

u

a

u

C

a

u

a

u

a

a

u

a

u

du

a

du

du

u

a

a

a

u

u

a

du

u

u

t

dt

udu

t

u

t

a

dt

t

t

a

x

xdx

dt

x

a

t

x

dx

x

a

x

x

dx

x

a

+

+

+

=

+

+

+

=

+

+

+

=

+

+

+

>=

<

=

+

=

+

=

=

=

=

=

=

=

=

=

=

=

=

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

ln

2

ln

2

ln

2

ln

2

1

51

2

2

1

2

62)

założenia:

0

2

2

>

x

a

(

)

(

)

C

x

a

x

a

x

a

C

a

x

a

x

a

a

x

a

x

x

x

C

a

x

a

x

a

a

x

a

C

t

t

a

t

a

C

t

a

t

a

C

u

a

t

a

udu

a

dt

a

dt

du

t

u

dt

t

a

dt

a

dt

t

a

x

x

tdt

a

tdt

a

x

a

dx

dt

a

x

t

t

a

x

x

a

dx

x

+

=

+

>=

=

=

=<

+

=

+

=

+

=

+

=

=

=

=

=

=

>=

=

=<

=

=

=

=

=

=

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

arcsin

2

1

2

arcsin

2

1

arcsin

cos

arcsin

sin

arcsin

cos

arcsin

sin

2

arcsin

2

cos

sin

2

2

2

sin

4

2

sin

4

2

cos

4

2

2

2

2

cos

2

2

2

2

cos

1

2

2

cos

1

sin

sin

sin

arcsin

sin

63)

(

)

C

x

a

C

t

dt

t

xdx

dt

x

a

t

dx

x

a

x

+

+

=

+

=

=

=

+

=

=

+

3

2

2

3

2

2

2

2

3

1

3

1

2

1

2

64)

(

)

(

)

(

)

=

+

>=

=

=<

+

=

+

=

+

=

=

=

=

+

+

=

+

dt

t

t

a

dt

t

a

x

x

tdt

a

dt

t

a

dt

t

a

a

t

a

x

a

dx

dt

a

x

t

t

a

x

x

a

dx

x

a

x

dx

x

a

x

1

cosh

sinh

sinh

1

cosh

sinh

sinh

sinh

sinh

sinh

arsinh

sinh

2

3

3

2

2

3

3

3

2

2

2

2

2

2

2

2

2

2

2



background image

(

)

(

)

(

)

C

x

a

C

a

x

a

x

x

C

a

x

a

C

t

a

C

u

a

C

u

a

u

a

t

a

du

u

a

tdt

a

tdt

du

t

u

+

+

=

+





+

>=

+

=

=<

+

=

+

=

+

=

+

+

=

+

=

=

=

3

2

2

2

2

2

3

2

3

3

3

3

3

3

3

3

3

3

2

3

3

3

1

1

3

1

arsinh

cosh

arsinh

cosh

3

cosh

3

3

3

cosh

1

sinh

sinh

cosh

65)

C

x

a

C

t

C

t

dt

t

t

dt

a

x

xdx

xdx

dt

a

x

t

a

x

xdx

+

+

=

+

=

+

=

=

=

+

=

=

+

=

=

+

2

2

2

1

2

1

2

1

2

2

2

2

2

2

2

1

2

1

2

1

2

1

2

2

1

2



66)

(

)

C

a

x

C

a

x

a

x

x

a

x

a

C

t

a

tdt

a

a

x

dx

dt

a

x

t

t

a

x

a

x

xdx

+

+

=

+

+

>=

+

=

=<

=

=

+

=

=

+

=

=

=

=

+

2

2

2

2

2

2

2

2

2

1

1

arsinh

cosh

arsinh

cosh

cosh

sinh

arsinh

sinh

67)

założenia:

0

x

C

x

C

t

t

dt

x

dx

dt

x

t

x

x

dx

x

x

x

dx

x

x

dx

x

dx

+

=

+

=

=

=

=

=

=

=

=

2

tgh

ln

ln

2

cosh

2

2

tgh

2

cosh

2

tgh

2

2

cosh

2

cosh

2

tgh

2

2

cosh

2

sinh

2

sinh

2

2

68)

C

cx

c

C

t

c

dt

t

c

cdx

dt

cx

t

cdx

cx

c

cxdx

+

=

+

=

=

=

=

=

=

cosh

1

cosh

1

sinh

1

sinh

1

sinh

69)

C

cx

c

C

t

c

tdt

c

cdx

dt

cx

t

cdx

cx

c

dx

cx

+

=

+

=

=

=

=

=

=

sinh

1

sinh

1

cosh

1

cosh

1

cosh

70)

C

x

x

dx

x

dx

x

x

x

xdx

+

>

=<

=

=

=

=

2

1

2

sinh

4

1

69

2

1

2

cosh

2

1

2

1

2

cosh

2

1

2

cosh

sinh

sinh

2

2

background image

71)

C

x

x

dx

dx

x

dx

x

x

x

xdx

+

+

>=

=<

+

=

+

=

+

=

=

2

1

2

sinh

4

1

69

2

1

2

cosh

2

1

2

1

2

cosh

2

1

2

cosh

cosh

cosh

2

2

72)

założenia:

1

;

0

;

0

>

>

a

a

x

(

)

C

a

x

x

x

C

a

x

a

x

x

C

x

x

a

xdx

a

a

xdx

xdx

a

a

+

=

+

=

+

=

=

=

=

ln

log

ln

ln

ln

1

ln

ln

1

4

ln

ln

1

ln

ln

log


73)

założenia:

1

;

0

;

0

>

>

a

a

x

C

a

x

x

x

dx

a

x

x

dx

a

x

x

x

x

x

v

a

x

du

dv

x

u

xdx

a

a

a

a

a

+

=

=

=

=

=

=

=

=

ln

log

ln

1

log

ln

log

ln

1

1

log

log

74)

C

x

C

x

C

t

t

dt

xdx

dt

x

t

dx

x

x

xdx

+

=

+

=

+

=

=

=

=

=

=

cosh

ln

cosh

ln

ln

sinh

cosh

cosh

sinh

tgh

75)

założenia:

0

x

C

x

C

t

t

dt

xdx

dt

x

t

dx

x

x

xdx

+

=

+

=

=

=

=

=

=

sinh

ln

ln

cosh

sinh

sinh

cosh

ctgh








Spis treści:

Funkcja

Całka

Założenia

Numer

Całki funkcji wymiernych

2

2

1

x

a

a

x

a

arctg

1

1

n

a

x

A

)

(



+

=

+

1

;

)

(

1

1

;

|

|

ln

1

n

C

a

x

n

A

n

C

a

x

A

n

0

a

x

5

b

ax

d

cx

+

+

|

|

ln

2

a

b

x

a

bc

ad

x

a

c

+

+

0

;

0

+

c

a

b

ax

6

n

b

ax

)

(

+




+

+

+

=

+

+

+

1

;

)

1

(

)

(

1

;

)

ln(

1

n

C

n

a

b

ax

n

C

a

b

ax

n

21

background image

n

q

px

x

p

x

)

(

2

2

+

+

+



+

+

+

=

+

+

+

1

;

1

)

(

1

;

|

|

ln

1

2

2

n

C

n

q

px

x

n

C

q

px

x

n

4

;

0

2

<

+

+

q

p

q

px

x

24

n

x

x

)

1

(

2

+

+

+

=

+

+

1

;

2

2

)

1

(

1

;

|

1

|

ln

2

1

1

2

2

n

C

n

x

n

C

x

n

25

n

x

)

1

(

1

2

+

+

+

+

1

2

1

2

)

1

(

2

2

3

2

)

1

(

2

2

1

n

n

x

dx

n

n

x

x

n

1

;

>

n

Z

n

26

n

q

px

x

dx

)

(

2

+

+

+

n

n

t

dt

)

1

(

)

4

(

2

2

1

+

<

Z

n

q

p

;

0

4

2

27

n

q

px

x

B

Ax

)

(

2

+

+

+

(

)

+

+

+

+

+

+

+

+

n

n

n

u

du

Ap

B

n

q

px

x

A

C

p

q

p

x

arctg

p

q

Ap

B

q

px

x

A

1

)

4

)(

2

(

)

1

(

2

)

(

4

2

)

2

4

)(

2

(

)

ln(

2

2

2

1

1

2

2

2

2

0

4

;

2

<

+

q

p

Z

n

28

b

k

x

+

2

)

(

1

b

k

x

arctg

b

1

+

R

b

29

(

)

n

b

ax

x

+

(

)

(

)(

)(

)

1

2

2

1

1

+

+

+

+

n

b

ax

n

n

a

b

x

n

a

{

}

1

,

2

n

40

b

ax

x

+

C

b

ax

a

b

a

b

a

x

+

+

+

ln

2

2

0

+

b

ax

41

(

)

2

b

ax

x

+

(

)

b

ax

a

b

ax

a

+

+

+

ln

1

1

2

2

0

+

b

ax

42

(

)

n

b

ax

x

+

( )

( )(

)(

)

1

2

2

1

1

+

n

b

ax

n

n

a

b

x

n

a

{ }

2

,

1

;

0

+

n

b

ax

43

b

ax

x

+

2

(

)

(

)

+

+

+

+

b

ax

b

b

ax

b

b

ax

a

ln

2

2

1

2

2

3

0

+

b

ax

44

(

)

2

2

b

ax

x

+

+

+

+

b

ax

b

b

ax

b

b

ax

a

2

3

ln

2

1

0

+

b

ax

45

(

)

3

2

b

ax

x

+

(

)

+

+

+

+

2

2

3

2

2

ln

1

b

ax

b

b

ax

b

b

ax

a

0

+

b

ax

46

(

)

n

b

ax

x

+

2

(

)(

)

(

)(

)

( )(

)

+

+

+

+

1

2

2

3

3

1

2

2

3

1

1

n

n

n

b

ax

n

b

b

ax

n

b

b

ax

n

a

{ }

3

,

2

,

1

;

0

+

n

b

ax

47

(

)

b

ax

x

+

1

x

b

ax

b

+

ln

1

(

)

0

;

0

+

b

b

ax

x

48

(

)

b

ax

x

+

2

1

x

b

ax

b

a

bx

+

+

ln

1

2

(

)

0

+

b

ax

abx

49

(

)

2

2

1

b

ax

x

+

(

)





+

+

+

x

b

ax

b

x

ab

b

ax

b

a

ln

2

1

1

3

2

2

(

)

0

+

b

ax

abx

50

background image

2

2

1

a

x

a

x

a

x

a

+

ln

2

1

0

2

2

a

x

51

2

2

1

a

x

(

)

(

) ( )

+

+

,

1

1

,

;

arctg

1

,

;

artgh

1

x

C

a

x

a

a

a

x

C

a

x

a

0

2

2

a

x

52

Całki funkcji niewymiernych

2

2

1

x

a

a

x

arcsin

0

2

2

>

x

a

9

2

2

1

x

a

a

x

arcsin

0

2

2

>

x

a

56

2

2

x

a

2

2

2

2

arcsin

2

x

a

x

a

x

a

+

0

2

2

>

x

a

10

2

2

x

a

2

2

2

2

arcsin

2

x

a

x

a

x

a

+

0

2

2

>

x

a

57

k

x

+

2

1

|

|

ln

2

k

x

x

+

+

0

>

k

11

2

2

1

a

x

+

a

x

arsinh

59

k

x

+

2

k

x

x

k

x

x</