Ad. 1
ńł �ł
x
�łu = ln |x| v �ł

=
�ł żł
x ln |x| dx ln |x| 1
(x2 + 1)2
I = = = - + dx
1 1
(x2 + 1)2 �łu = 1 �ł 2(x2 + 1) 2x(x2 + 1)

ół �ł
v = -
x 2 x2 + 1
1 (1 + x2) - x2 1 x
= = -
2x(x2 + 1) 2x(x2 + 1) 2x 2(x2 + 1)
ln |x| 1 1 x2 ln |x| 1
I = - + ln |x| - ln(x2 + 1) + C = - ln(x2 + 1) + C
2(x2 + 1) 2 4 2(x2 + 1) 4
ńł �ł
�łu = ln |x| v 1 �ł

�ł żł
=
ln |x| dx ln |x| 1
(x + 2)2
I = = = - + dx =
1
(x + 2)2 �łu 1 �ł x + 2 x(x + 2)
ół �ł
= v = -
x x + 2



ln |x| 1 1 1 ln |x| 1 1 ln |x| 1 x

= - + - dx = - + ln |x| - ln |x + 2| + C = - + ln
x + C
x + 2 2 x x + 2 x + 2 2 2 x + 2 2 + 2

1 1 1 x ln |x| 1
lub: = - + ln |x| - ln |x + 2| + C = - ln |x + 2| + C
x + 2 2 2 2(x + 2) 2
ńł �ł
"
"
�łu = arc tg x v = 1żł
" "
x
1 1
I = arc tg x dx == = x arc tg x - dx =
u = � " v = x
ół �ł 2(1 + x)
1 + x 2 x

" "
"
t � 2t dt t2 dt
= x = t, x = t2, dx = 2t dt = x arc tg x - = x arc tg x -
2(1 + t2) (1 + t2)
t2 t2 + 1 - 1 1
Ponieważ = = 1 -
(1 + t2) t2 + 1 t2 + 1
" " " " " "
więc I = x arc tg x - t + arc tg t + C = arc tg x - x + arc tg x = (x + 1) arc tg x - x + C


u = ln(x2 + x + 2) v = x2
1 1 2x4 + x3
2x + 1 1
I = x2 ln(x2+x+2) dx = = x3 ln(x2+x+2)- dx
u = v = x3
3 3 x2 + x + 2
x2 + x + 2 3
Dzielenie daje:
2x2 - x - 3

x2 + x + 2 2x4 + x3
- 2x4 - 2x3 - 4x2
- x3 - 4x2
x3 + x2 + 2x
- 3x2 + 2x
3x2 + 3x + 6
5x + 6


5 7
(2x + 1) +
2x4 + x3 5x + 6 2 1
2 2
dx = 2x2 - x - 3 + dx = x3 - x2 - 3x + dx =
x2 + x + 2 x2 + x + 2 3 2 x2 + x + 1

"
2 1 5 7 dx 2 1 5 2x + 1
= x3- x2-3x+ ln(x2+x+2)+ = x3- x2-3x+ ln(x2+x+2)+ 7 arc tg " +C
2 7
1
3 2 2 2 3 2 2
7
x + +
2 4
"
1 2 1 5 7 2x + 1
więc I = x3 ln(x2 + x + 2) - x3 + x2 + x - ln(x2 + x + 2) - arc tg " + C =
3 9 6 6 3
7
"

1 5 x(4x2 - 3x - 18) 7 2x + 1
= x3 - ln(x2 + x + 2) - - arc tg " + C
3 6 18 3
7
ńł �ł

�łu = (arc sin x)3 v = 1żł
3x
1
(arc sin x)3 dx = = x(arc sin x)3- " (arc sin x)2 dx =
u = 3(arc sin x)2 � " v = x
ół �ł
1 - x2
1 - x2
ńł �ł
3 � (2x)
�łu = (arc sin x)2 �ł

�ł v = " żł
"
2 1 - x2
= = x(arc sin x)3+3 1 - x2(arc sin x)2- 6 arc sin x dx =
"
�łu 2(arc sin x) � 1
ół �ł
= " v = -3 1 - x2�ł
1 - x2
ńł �ł
�łu = 6 arc sin x v = 1żł
" "
6
= = x(arc sin x)3 + 3 1 - x2(arc sin x)2 - 6x arc sin x - 6 1 - x2 + C
u = " v = x
ół �ł
1 - x2

x3ex
I = dx
(x + 3)2
Dzielenie x3 przez (x + 3)2 = x2 + 6x + 9 daje
x - 6

x2 + 6x + 9 x3
- x3 - 6x2 - 9x
- 6x2 - 9x
6x2 + 36x + 54
27x + 54
więc


(27x + 54)ex
I = (x - 6)ex + dx;
(x + 3)2


(27x + 54)ex (27x + 84 - 27)ex 27 27
Tutaj dx = dx = - ex dx
(x + 3)2 (x + 3)2 x + 3 (x + 3)2

27 27
= ex dx - ex dx
x + 3 (x + 3)2
ńł �ł
�łu = 27

�ł żł
v = ex�ł 27 27
27
x + 3
= = ex + ex dx - ex dx =

�łu = - 27 �ł (x + 3)2 (x + 3)2
ół v = ex x + 3
�ł
(x + 3)2
27
= ex + C
x + 3

27 27
Więc I = xex - ex - 6ex + ex + C = + x - 7 ex + C
x + 3 x + 3
Ad. 2
ńłx + 1 �ł
1
�ł �ł
�ł = t2 x = �ł

�ł �ł

�ł żł
x t2 - 1
1 x + 1 2t
-2t
dx = = - (t2-1)2t dt = - 2t2 dt =
dt
x2 x �łx + 1 = xt2 dx = (t2 - 1)2
�ł �ł
(t2 - 1)2 �ł
�ł �ł
ół �ł
x(1 - t2) = -1
3/2
2 2 x + 1
= - t3 + C = - + C
3 3 x
x + 1 1 1
Ewentualnie, podstawienie u = = 1 + , du = - dx.
x x x2
ńłx - 2 �ł
2
�ł �ł
�ł = t2 x = - �ł

�ł �ł

�ł żł
x t2 - 1
1 x - 2 t2 - 1 4t 2t2
4t
dx = = - t dt = - dt =
dt
x x �łx - 2 = xt2 dx = 2 (t2 - 1)2 t2 - 1
�ł �ł
(t2 - 1)2 �ł
�ł �ł
ół �ł
x(1 - t2) = 2


t - 1
(2t2 - 2) + 2 1 1

= - dt = - 2 + - dt = - 2t + ln
t + 1 + C =
t2 - 1 t - 1 t + 1

2


x-2
x-2
- 1 - 1

x - 2 x - 2 x
x
x-2
= -2 - ln + C = -2 - ln =


x x-2 x
- 1

+ 1 x
x



x - 2 x - 2 x - 2 x - 2


= -2 - 2 ln - 1 + ln 2 - ln |x| + C = -2 - 2 ln - 1 - ln |x| + C

x x x x

"
dx 2t dt 2
" = ( x - 1 = t, x - 1 = t2, x = t2 + 1, dx = 2t dt) = = dt =
t2 - t t - 1
x - 1 - x - 1
"

= 2 ln |t - 1| + C = 2 ln x - 1 - 1 + C

" " "
I = arc tg 1 + x dx = (1 + x = t, x = t - 1, x = (t - 1)2, dx = 2(t - 1) dt)


u = arc tg t v = 2t - 2
t2 - 2t
= arc tg t � 2(t - 1) dt = 1 = (t2 - 2t) arc tg t - dt
u = v = t2 - 2t
t2 + 1
1 + t2
t2 - 2t t2 + 1 - 2t - 1 2t + 1
= = 1 -
t2 + 1 t2 + 1 t2 + 1
I = (t2 - 2t) arc tg t - t + ln(t2 + 1) + arc tg t + C
" " " "
a ponieważ t2 - 2t = 1 + 2 x + x - 2 - 2 x = x - 1 i t2 + 1 = x + 2 x + 1 + 1 = t2 + 2 x + 2
" " " "
więc I = (x - 1) arc tg 1 + x - 1 - x + ln x + 2 x + 2 + arc tg 1 + x =
" " "


= x arc tg 1 + x - x + ln x + 2 x + 2 + C
Ad. 4

dx
I =
sin4 x + cos4 x
1
sin4 x + cos4 x = (sin2 x + cos2 x)2 - 2 sin2 x cos2 x = 1 - sin2 x cos2 x;
2

dx 2 dx
I = = =
1
1 - sin2 2x 2 - sin2 2x
2

2t 1
= tg x = t; sin 2x = ; x = arc tg t + kĄ; dx = dt =
1 + t2 1 + t2

1 1 1 + t2 t2 + 1
= � dt = dt = dt
4t2
1 + t2 2(1 + t2)2 - 4t2 2(t4 + 1)
2 -
1+t2
t2 + 1 At + B Ct + D
= " + "
2(t4 + 1)
t2 - 2t + 1 t2 + 2t + 1
" "
t2 + 1 = 2(At + B)(t2 + 2t + 1) + 2(Ct + D)(t2 - 2t + 1)
" "
t2 + 1 = 2At3+2Bt2+2A 2t2+2B 2t+2At+2B+
" "
+ 2Ct3+2Dt2-2C 2t2-2D 2t+2Ct+2D
t3 : 2A + 2C = 0
" "
t2 : 2A 2+2B - 2C 2 + 2D = 1
" "
t : 2A +2B 2 + 2C - 2D 2 = 0
1 : 2B + 2D = 1
1
Z pierwszego i trzeciego równania B = D, więc na mocy czwartego B = D = .
4
Teraz z drugiego A - C = 0, co w połączeniu z pierwszym daje A = C = 0. Zatem
" "


1 1 1 2 2t + 2 1 2 2t - 2
" "
I = " + " dt = arc tg " + arc tg " + C =
4 4
4(t2 - 2t + 1) 4(t2 + 2t + 1) 2 2 2 2
" "
1 1
" "
= arc tg 2t + 1 + arc tg 2t - 1 + C =
2 2 2 2
" "
1 1
" "
= arc tg 2 tg x + 1 + arc tg 2 tg x - 1 + C
2 2 2 2
Ad. 7


dx x 2 2t 1 - t2
= t = tg ; x = 2 arc tg t + 2kĄ; dx = dt; sin x = ; cos x = ;
1 + sin x + cos x 2 1 + t2 1 + t2 1 + t2


2
dt
x 1 - cos x sin x 2 2
1+t2
tg = = = = dt = dt =
2t 1-t2
2 sin x 1 + cos x 1 + t2 + 2t + 1 - t2 2t + 2
1 + +
1+t2 1+t2



1 - cos x sin x - cos x + 1
1
tg x

= dt = ln |t + 1| + C = ln + 1 + C = ln + 1 + C = ln + C


t + 1 2 sin x sin x

sin x + cos x + 1
sin x

lub: = ln + C
1 + cos x + 1 + C = ln

1 + cos x