Question

Integrate:

$\int \frac{1-x}{2x^2+1}dx$

Answer 1

$\int \frac{1-x}{2x^2+1}dx=\int \frac{1}{2x^2+1}dx-\int \frac{xdx}{2x^2+1}$

$=\frac{1}{2}\int \frac{1}{x^2+\frac{1}{2}}dx-\frac{1}{4}\int \frac{4xdx}{2x^2+1}$

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

$=\frac{1}{2}.\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}{\tan }^{-1}\left(\frac{x}{\left(\frac{1}{\sqrt{2}}\right)}\right)-\frac{1}{4}\log (2x^2+1)+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}x\right)-\frac{1}{4}\log (2x^2+1)+C$