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

Question

Aicharya · Accepted Answer

Let $\begin{aligned} I &=\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x \ &=\int \frac{x^{2}\left(1+\frac{1}{x^{2}}ight)}{x^{2}\left(x^{2}+1+\frac{1}{x^{2}}ight)} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}-2ight)+3} \ I &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}ight)^{2}+(\sqrt{3})^{2}} d x \end{aligned}$ Put $x-\frac{1}{x}=t \Rightarrow\left(1+\frac{1}{x^{2}}ight) d x=d t$ $	herefore I=\int \frac{d t}{t^{2}+(\sqrt{3})^{2}}$ $=\frac{1}{\sqrt{3}} 	an ^{-1}\left(\frac{t}{\sqrt{3}}ight)+c=\frac{1}{\sqrt{3}} 	an ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}ight)+c$

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

Solution