Question

Вычислить неопределенный интеграл

Answer 1

Ответ:

решение на фотографиях

Answer 2

$1)\; \; \int \dfrac{e^{x}+sinx}{(e^{x}-cosx)^2}\, dx=\Big[\; u=e^{x}-cosx\; ,\; du=(e^{x}+sinx)dx\; \Big]=\\\\=\int \dfrac{du}{u^2}=-\dfrac{1}{u}+C=-\dfrac{1}{e^{x}-cosx}+C=\dfrac{1}{cosx-e^{x}}+C$

$2)\; \; \int \dfrac{cosx\, dx}{\sqrt{2+cos2x}}=\Big[\; cos2x=1-2sin^2x\; \Big]=\int \dfrac{cosx\, dx}{\sqrt{3-2sin^2x}}=\\\\=\Big[\; u=\sqrt2sinx\; ,\; du=\sqrt2\, sinx\, dx\; \Big]=\dfrac{1}{\sqrt2}\int \dfrac{du}{\sqrt{3-u^2}}=\\\\=\dfrac{1}{\sqrt2}\cdot arcsin\frac{u}{\sqrt3}+C=\dfrac{1}{\sqrt2}\cdot arcsin\frac{\sqrt2\, sinx}{\sqrt3}+C$

$3)\; \; \int \dfrac{dx}{x\sqrt{1+x^2}}=\Big[\; x=\frac{1}{t}\; ,\; dx=-\frac{dt}{t^2}\; \Big]=\int \dfrac{-\frac{dt}{t^2}}{\frac{1}{t}\sqrt{1+\frac{1}{t^2}}}=-\int \dfrac{dt}{t\, \sqrt{\frac{1+t^2}{t^2}}}=\\\\=-\int \frac{dt}{\sqrt{1+t^2}}=-ln\Big|t+\sqrt{1+t^2}\Big|+C=\\\\=-ln\Big|\frac{1}{x}+\frac{\sqrt{1+x^2}}{x}\Big|+C=-ln\Big|\frac{1+\sqrt{1+x^2}}{x}\Big|+C$

$4)\; \; \int \dfrac{2^{x}\, dx}{\sqrt{1-4^{x}}}=\dfrac{1}{ln2} \int \dfrac{d(2^{x})}{\sqrt{1-(2^{x})^2}}=\dfrac{1}{ln2}\cdot arcsin(2^{x})+C$

$5)\; \; \int sin\sqrt{x}\, dx=\Big[\; x=t^2\; ,\; dx=2t\, dt\; ,\; \sqrt{x}=t\; \Big]=\int sint\cdot 2t\, dt=\\\\=\Big[\; u=t\; ,\; du=dt\; ,\; v=\int sint\, dt=-cost\; \Big]=2\cdot \Big(-t\cdot cost+\int cost\, dt\Big)=\\\\=-2t\, cost+2\, sint+C=-2\sqrt{x}\cdot cos\sqrt{x}+2\, sin\sqrt{x}+C$

$6)\; \; \int \dfrac{dx}{\sqrt{1+e^{2x}}}=\Big[\; 1+e^{2x}=t^2\; ,\; e^{2x}=t^2-1\; ,\; 2x=ln(t^2-1)\; ,\\\\x=\frac{1}{2}ln(t^2-1)\; ,\; dx=\dfrac{2t\, dt}{2(t^2-1)}\; \Big]=\int \dfrac{t\, dt}{(t^2-1)\cdot t}=\int \dfrac{dt}{t^2-1}=\\\\\\=\frac{1}{2}\cdot ln\Big| \dfrac{t-1}{t+1}\Big|+C=\dfrac{1}{2}\cdot ln\Big|\dfrac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\Big|+C$