Question

Решить интегралы на фото

Answer 1

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Answer 2

$1)\; \; \int \frac{x+5}{x^3}dx=\int (\frac{1}{x^2}+\frac{5}{x^3})dx=\frac{x^{-1}}{-1}+\frac{5x^{-2}}{-2}+C=-\frac{1}{x}-\frac{5}{2x^2}+C\\\\2)\; \; \int (sin2x-cos2x)dx=-\frac{1}{2}cos2x-\frac{1}{2}sin2x+C\\\\3)\; \; \int \frac{d}{\sqrt{11-x^2}}=arcsin\frac{x}{\sqrt{11}}+C\\\\4)\; \; \int \frac{e^{x}-2}{e^{2x}+4}dx=\int \frac{e^{x}dx}{(e^{x})^2+2^2}-2\int \frac{dx}{e^{2x}+4}=\int \frac{d(e^{x})}{(e^{x})^2+2^2}-2\Big [\, t=e^{2x} ,\\\\2x=lnt,\; x=\frac{1}{2}lnt,\; dx=\frac{dt}{2t}\, ]=\frac{1}{2}arctg\frac{e^{x}}{2}-$

$-2\int \frac{dt}{2t(t+4)}=\frac{1}{2}arctg\frac{e^{x}}{2} -\int \frac{dt}{t^2+4t}=\frac{1}{2}arctg\frac{e^{x}}{2}-\int \frac{dt}{(t+2)^2-4}=\\\\=\frac{1}{2}arctg\frac{e^{x}}{2}-\frac{1}{4}\cdot ln\Big |\frac{t+2-2}{t+2+2}\Big |+C=\frac{1}{2}arctg\frac{e^{x}}{2}-\frac{1}{4}\cdot ln\Big |\frac{e^{2x}}{e^{2x}+4}\Big |+C$

$5)\; \; \int x^2\, e^{2x}dx=[\, u=x^2,\; du=2xdx,dv=e^{2x}dx,\; v=\frac{1}{2}e^{2x}\, ]=\\\\=\frac{x^2}{2}e^{2x}-\int x\, e^{2x}dx=[\, u=x,\; du=dx,\; v=\frac{1}{2}e^{2x}\, ]=\\\\=\frac{x^2}{2}e^{2x}-(\frac{x}{2}e^{2x}-\frac{1}{2} \int e^{2x}dx)=\frac{x^2}{2}e^{2x}-\frac{x}{2}e^{2x}+\frac{1}{4}e^{2x}+C=\\\\=\frac{e^{2x}}{4}\cdot (2x^2-2x+1)+C$

$6)\; \; \int \frac{sin3x}{4+cos^23x}dx=\frac{1}{3}\int \frac{-d(cos3x)}{4+cos^23x}=-\frac{1}{3}\cdot \frac{1}{2}\cdot arctg\frac{cos3x}{2}+C\\\\7)\; \; \int \frac{\sqrt{1+x}}{x}dx=[\, 1+x=t^2,\; t=\sqrt{1+x},\; x=t^2-1,\; dx=2tdt]=\\\\=\int \frac{t\cdot 2t\, dt}{t^2-1}=2\int\frac{t^2\, dt}{t^2-1}=2\int (1+\frac{1}{t^2-1})dt=2(t+\frac{1}{2}\cdot ln\Big |\frac{t-1}{t+1}\Big |)+C=\\\\=2\sqrt{1+x}+ln\Big |\frac{\sqrt{1+x}\, -1}{\sqrt{1+x}\, +1}\Big |+C$