Question

Помогите с решением интегралов, пожалуйста

Answer 1

$\displaystyle \int \frac{cos2xdx}{sin^2xcos^2x}=2\int\frac{cos2xd(2x)}{sin^22x}=2\int\frac{d(sin2x)}{sin^22x}=-\frac{2}{sin2x}+C$

$\displaystyle \int\frac{x^4dx}{\sqrt{4+x^5}}=\frac{1}{5}\int\frac{d(4+x^5)}{\sqrt{4+x^5}}=\frac{2}{5}\sqrt{4+x^5}+C$

$\displaystyle \int cos^2xsin2xdx=2\int cos^3xsinxdx=-2\int cos^3xd(cosx)=\\=-\frac{cos^4x}{2}+C$

$\displaystyle \int ln^2xdx=xln^2x-2\int lnxdx=xln^2x-2xlnx+2\int dx=\\=xln^2x-2xlnx+2x+C\\\\\\\u=ln^2x\rightarrow du=\frac{2lnx}{x}dx\\dv=dx\rightarrow v=x\\\\p=lnx\rightarrow dp=\frac{dx}{x}\\dq=dx\rightarrow q=x$

$\displaystyle \int xe^{-2x}dx=-\frac{x}{2}e^{-2x}+\frac{1}{2}\int e^{-2x}dx=-\frac{x}{2}e^{-2x}-\frac{1}{4}e^{-2x}+C\\u=x\rightarrow du=dx\\dv=e^{-2x}dx\rightarrow v=-\frac{1}{2}e^{-2x}$

$\displaystyle \int\frac{xdx}{x^2-8x+9}=\frac{1}{2}\int\frac{(2x-8+8)dx}{x^2-8x+9}=\frac{1}{2}\int\frac{d(x^2-8x+9)}{x^2-8x+9}+\\+4\int\frac{dx}{x^2-8x+9}=\frac{1}{2}\int\frac{d(x^2-8x+9)}{x^2-8x+9}+4\int\frac{d(x-4)}{(x-4)^2-7}=\\=\frac{1}{2}ln|x^2-8x+9|+\frac{2}{\sqrt7}ln|\frac{x-4-\sqrt7}{x-4+\sqrt7}|+C=\\=\frac{1}{2}ln|x-4-\sqrt7|+\frac{1}{2}ln|x-4+\sqrt7|+\frac{2}{\sqrt{7}}ln|x-4-\sqrt7|-\\-\frac{2}{\sqrt 7}ln|x-4+\sqrt7|+C=(\frac{7+4\sqrt7}{14})ln|x-4-\sqrt7|+\\+(\frac{7-4\sqrt7}{14})ln|x-4+\sqrt7|+C$

$(x^2-8x+9)'=2x-8$