∫dx = x + c
$$\int_{}^{}{x^{n}\ dx = \frac{x^{n + 1}}{n + 1}} + c$$
$$\int_{}^{}{\frac{1}{x}\ dx = \ln\left| x \right|} + c$$
$$\int_{}^{}{a^{x}\ dx = \frac{a^{x}}{\ln a}} + c$$
∫ex dx = ex + c
$$\int_{}^{}{e^{\text{ax}}\ dx =}\frac{1}{a}e^{\text{ax}} + c$$
∫sinax dx = -cosax + c
∫cosax dx = sinax + c
∫tgx dx = −ln|cosx| + c
∫ctgx dx=ln|sinx| + c
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{-ctg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{tg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} + a^{2}} =}\frac{1}{a}\operatorname{arc\ tg}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} - a^{2}} =}\frac{1}{2a}\ln\left| \left. \ \frac{x - a}{x + a} \right| \right.\ + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{a^{2} - x^{2}}} =}\operatorname{arc\ sin}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{x^{2} + q}} =}\ln\left| x + \right.\ \left. \ \sqrt{x^{2} + q} \right| + c$$
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)]dx=∫f(x)dx-∫g(x)dx
∫af(x)dx = a∫f(x)dx
cos2x = cos2x − sin2x
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(C)’ = 0
(xn)′ = nxn − 1
(x)’ = 1
($\frac{a}{x})' = \ - \ \frac{a}{x^{2}}$
($\sqrt{x)}$’ = $\frac{1}{2\sqrt{x}}$
(ax)′ = axlna
(ex)′ = ex
$\left( \log_{a}x \right)^{'} = \ \frac{1}{\text{xlna}}$
$\left( \text{lnx} \right)^{'} = \ \frac{1}{x}$
(sinx)′ = cosx
(cosx)’ = -sinx
(tgx)’ = $\frac{1}{\cos^{2}x}$
(ctgx)’ = - $\frac{1}{{- \sin}^{2}x}$
(arcsinx)’ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arccos)’ = -$\frac{1}{\sqrt{1 - x^{2}}}$
(arctgx)’ = $\frac{1}{x^{2} + 1}$
(arcctgx)’ = - $\frac{1}{x^{2} + 1}$
[F(x)*g(x)]’=f’(x)g(x)+f(x)g’(x)
$$(\sqrt{})' = \frac{1}{2\sqrt{}}*^{2}$$
(2)′ = 2 * *’
∫dx = x + c
$$\int_{}^{}{x^{n}\ dx = \frac{x^{n + 1}}{n + 1}} + c$$
$$\int_{}^{}{\frac{1}{x}\ dx = \ln\left| x \right|} + c$$
$$\int_{}^{}{a^{x}\ dx = \frac{a^{x}}{\ln a}} + c$$
∫ex dx = ex + c
$$\int_{}^{}{e^{\text{ax}}\ dx =}\frac{1}{a}e^{\text{ax}} + c$$
∫sinax dx = -cosax + c
∫cosax dx = sinax + c
∫tgx dx = −ln|cosx| + c
∫ctgx dx=ln|sinx| + c
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{-ctg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{tg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} + a^{2}} =}\frac{1}{a}\operatorname{arc\ tg}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} - a^{2}} =}\frac{1}{2a}\ln\left| \left. \ \frac{x - a}{x + a} \right| \right.\ + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{a^{2} - x^{2}}} =}\operatorname{arc\ sin}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{x^{2} + q}} =}\ln\left| x + \right.\ \left. \ \sqrt{x^{2} + q} \right| + c$$
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)]dx=∫f(x)dx-∫g(x)dx
∫af(x)dx = a∫f(x)dx
cos2x = cos2x − sin2x
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(C)’ = 0
(xn)′ = nxn − 1
(x)’ = 1
($\frac{a}{x})' = \ - \ \frac{a}{x^{2}}$
($\sqrt{x)}$’ = $\frac{1}{2\sqrt{x}}$
(ax)′ = axlna
(ex)′ = ex
$\left( \log_{a}x \right)^{'} = \ \frac{1}{\text{xlna}}$
$\left( \text{lnx} \right)^{'} = \ \frac{1}{x}$
(sinx)′ = cosx
(cosx)’ = -sinx
(tgx)’ = $\frac{1}{\cos^{2}x}$
(ctgx)’ = - $\frac{1}{{- \sin}^{2}x}$
(arcsinx)’ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arccos)’ = -$\frac{1}{\sqrt{1 - x^{2}}}$
(arctgx)’ = $\frac{1}{x^{2} + 1}$
(arcctgx)’ = - $\frac{1}{x^{2} + 1}$
[F(x)*g(x)]’=f’(x)g(x)+f(x)g’(x)
$$(\sqrt{})' = \frac{1}{2\sqrt{}}*^{2}$$
(2)′ = 2 * *’
∫dx = x + c
$$\int_{}^{}{x^{n}\ dx = \frac{x^{n + 1}}{n + 1}} + c$$
$$\int_{}^{}{\frac{1}{x}\ dx = \ln\left| x \right|} + c$$
$$\int_{}^{}{a^{x}\ dx = \frac{a^{x}}{\ln a}} + c$$
∫ex dx = ex + c
$$\int_{}^{}{e^{\text{ax}}\ dx =}\frac{1}{a}e^{\text{ax}} + c$$
∫sinax dx = -cosax + c
∫cosax dx = sinax + c
∫tgx dx = −ln|cosx| + c
∫ctgx dx=ln|sinx| + c
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{-ctg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{tg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} + a^{2}} =}\frac{1}{a}\operatorname{arc\ tg}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} - a^{2}} =}\frac{1}{2a}\ln\left| \left. \ \frac{x - a}{x + a} \right| \right.\ + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{a^{2} - x^{2}}} =}\operatorname{arc\ sin}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{x^{2} + q}} =}\ln\left| x + \right.\ \left. \ \sqrt{x^{2} + q} \right| + c$$
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)]dx=∫f(x)dx-∫g(x)dx
∫af(x)dx = a∫f(x)dx
cos2x = cos2x − sin2x
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(C)’ = 0
(xn)′ = nxn − 1
(x)’ = 1
($\frac{a}{x})' = \ - \ \frac{a}{x^{2}}$
($\sqrt{x)}$’ = $\frac{1}{2\sqrt{x}}$
(ax)′ = axlna
(ex)′ = ex
$\left( \log_{a}x \right)^{'} = \ \frac{1}{\text{xlna}}$
$\left( \text{lnx} \right)^{'} = \ \frac{1}{x}$
(sinx)′ = cosx
(cosx)’ = -sinx
(tgx)’ = $\frac{1}{\cos^{2}x}$
(ctgx)’ = - $\frac{1}{{- \sin}^{2}x}$
(arcsinx)’ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arccos)’ = -$\frac{1}{\sqrt{1 - x^{2}}}$
(arctgx)’ = $\frac{1}{x^{2} + 1}$
(arcctgx)’ = - $\frac{1}{x^{2} + 1}$
[F(x)*g(x)]’=f’(x)g(x)+f(x)g’(x)
$$(\sqrt{})' = \frac{1}{2\sqrt{}}*^{2}$$
(2)′ = 2 * *’
∫dx = x + c
$$\int_{}^{}{x^{n}\ dx = \frac{x^{n + 1}}{n + 1}} + c$$
$$\int_{}^{}{\frac{1}{x}\ dx = \ln\left| x \right|} + c$$
$$\int_{}^{}{a^{x}\ dx = \frac{a^{x}}{\ln a}} + c$$
∫ex dx = ex + c
$$\int_{}^{}{e^{\text{ax}}\ dx =}\frac{1}{a}e^{\text{ax}} + c$$
∫sinax dx = -cosax + c
∫cosax dx = sinax + c
∫tgx dx = −ln|cosx| + c
∫ctgx dx=ln|sinx| + c
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{-ctg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{tg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} + a^{2}} =}\frac{1}{a}\operatorname{arc\ tg}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} - a^{2}} =}\frac{1}{2a}\ln\left| \left. \ \frac{x - a}{x + a} \right| \right.\ + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{a^{2} - x^{2}}} =}\operatorname{arc\ sin}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{x^{2} + q}} =}\ln\left| x + \right.\ \left. \ \sqrt{x^{2} + q} \right| + c$$
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)]dx=∫f(x)dx-∫g(x)dx
∫af(x)dx = a∫f(x)dx
cos2x = cos2x − sin2x
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(C)’ = 0
(xn)′ = nxn − 1
(x)’ = 1
($\frac{a}{x})' = \ - \ \frac{a}{x^{2}}$
($\sqrt{x)}$’ = $\frac{1}{2\sqrt{x}}$
(ax)′ = axlna
(ex)′ = ex
$\left( \log_{a}x \right)^{'} = \ \frac{1}{\text{xlna}}$
$\left( \text{lnx} \right)^{'} = \ \frac{1}{x}$
(sinx)′ = cosx
(cosx)’ = -sinx
(tgx)’ = $\frac{1}{\cos^{2}x}$
(ctgx)’ = - $\frac{1}{{- \sin}^{2}x}$
(arcsinx)’ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arccos)’ = -$\frac{1}{\sqrt{1 - x^{2}}}$
(arctgx)’ = $\frac{1}{x^{2} + 1}$
(arcctgx)’ = - $\frac{1}{x^{2} + 1}$
[F(x)*g(x)]’=f’(x)g(x)+f(x)g’(x)
$$(\sqrt{})' = \frac{1}{2\sqrt{}}*^{2}$$
(2)′ = 2 * *’
∫dx = x + c
$$\int_{}^{}{x^{n}\ dx = \frac{x^{n + 1}}{n + 1}} + c$$
$$\int_{}^{}{\frac{1}{x}\ dx = \ln\left| x \right|} + c$$
$$\int_{}^{}{a^{x}\ dx = \frac{a^{x}}{\ln a}} + c$$
∫ex dx = ex + c
$$\int_{}^{}{e^{\text{ax}}\ dx =}\frac{1}{a}e^{\text{ax}} + c$$
∫sinax dx = -cosax + c
∫cosax dx = sinax + c
∫tgx dx = −ln|cosx| + c
∫ctgx dx=ln|sinx| + c
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{-ctg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\operatorname{}x}\ = \operatorname{tg}x} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} + a^{2}} =}\frac{1}{a}\operatorname{arc\ tg}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{x^{2} - a^{2}} =}\frac{1}{2a}\ln\left| \left. \ \frac{x - a}{x + a} \right| \right.\ + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{a^{2} - x^{2}}} =}\operatorname{arc\ sin}\frac{x}{a} + c$$
$$\int_{}^{}{\frac{\text{dx}}{\sqrt{x^{2} + q}} =}\ln\left| x + \right.\ \left. \ \sqrt{x^{2} + q} \right| + c$$
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)]dx=∫f(x)dx-∫g(x)dx
∫af(x)dx = a∫f(x)dx
cos2x = cos2x − sin2x
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(C)’ = 0
(xn)′ = nxn − 1
(x)’ = 1
($\frac{a}{x})' = \ - \ \frac{a}{x^{2}}$
($\sqrt{x)}$’ = $\frac{1}{2\sqrt{x}}$
(ax)′ = axlna
(ex)′ = ex
$\left( \log_{a}x \right)^{'} = \ \frac{1}{\text{xlna}}$
$\left( \text{lnx} \right)^{'} = \ \frac{1}{x}$
(sinx)′ = cosx
(cosx)’ = -sinx
(tgx)’ = $\frac{1}{\cos^{2}x}$
(ctgx)’ = - $\frac{1}{{- \sin}^{2}x}$
(arcsinx)’ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arccos)’ = -$\frac{1}{\sqrt{1 - x^{2}}}$
(arctgx)’ = $\frac{1}{x^{2} + 1}$
(arcctgx)’ = - $\frac{1}{x^{2} + 1}$
[F(x)*g(x)]’=f’(x)g(x)+f(x)g’(x)
$$(\sqrt{})' = \frac{1}{2\sqrt{}}*^{2}$$
(2)′ = 2 * *’