Интегриране по части

[west234]

[tex]\int[/tex] dx/(2+x^2)^2

[Nathi123]

I=[tex]\int x^{2}arccosdx=\frac{1}{3} \int arccosx dxx^{3}=\frac{x^{3}arccosx}{3}+\frac{1}{3}\int \frac{x^{3}dx}{\sqrt{1-x^{2}}}[/tex]
I=[tex]\frac{x^{3}arccosx}{3} - \frac{1}{3}\int \frac{x^{2}d(1-x^{2})}{ 2 \sqrt{1-x^{2}}}=\frac{x^{3}arccosx}{3} - \frac{1}{3}\int x^{2}d\sqrt{1-x^{2}}[/tex]
I= [tex]\frac{x^{3}arccosx}{3} -\frac{x^{2}\sqrt{1-x^{2}}}{3} + \frac{1}{3}\int 2x\sqrt{1-x^{2}dx}[/tex]
I = [tex]\frac{x^{3}arccosx}{3} -\frac{x^{2}\sqrt{1-x^{2}}}{3} - \frac{1}{3}\int \sqrt{1-x^{2}}d(1-x^{2})=\frac{x^{3}arccosx}{3} -\frac{x^{2}\sqrt{1-x^{2}}}{3} - \frac{2}{9}\sqrt{(1-x^{2})^{3}} + C.[/tex]

[Добромир Глухаров]

$\int\frac{dx}{(2+x^2)^2}=\frac{1}{2}\int\frac{2+x^2-x^2}{(2+x^2)^2}dx=\frac{1}{2}\int\frac{dx}{2+x^2}-\frac{1}{4}\int\frac{xd(2+x^2)}{(2+x^2)^2}=\\=\frac{1}{2}\int\frac{dx}{2+x^2}-\frac{1}{4}\int xd\left(\frac{(2+x^2)^{-1}}{-1}\right)=\frac{1}{2}\int\frac{dx}{2+x^2}+\frac{x}{4(2+x^2)}-\frac{1}{4}\int\frac{dx}{2+x^2}=\\=\frac{1}{4}\int\frac{dx}{2+x^2}+\frac{x}{4(2+x^2)}=\frac{1}{4\sqrt{2}}\int\frac{d\left(\frac{x}{\sqrt{2}}\right)}{1+\left(\frac{x}{\sqrt{2}}\right)^2}+\frac{x}{4(2+x^2)}=\frac{1}{4\sqrt{2}}arctg\left(\frac{x}{\sqrt{2}}\right)+\frac{x}{4(2+x^2)}+C$