Question

Evaluate : $\int \frac{x^2+1}{x^4+x^2+1}dx$

Answer 1

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

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

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

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

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

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

$=\frac{1}{2}\int \frac{1}{x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}}dx+\frac{1}{2}\int \frac{1}{x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}}dx$

$=\frac{1}{2}\int \frac{1}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx+\frac{1}{2}\int \frac{1}{{(x+\frac{1}{x})}^2+\frac{3}{4}}dx$

$=\frac{1}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x-1/2}{\sqrt{3}/2}+\frac{1}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x+1/2}{\sqrt{3}/2}+C$

$=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$