Question

$\int \frac{x^2+1}{x^4+x^2+1}dx$ is equal to (where $C$ is integration constant)

• A. ${\tan }^{-1}\left(\frac{x^2-1}{x}\right)+C$
• B. $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+C$
• C. $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2+1}{\sqrt{3}x}\right)+C$
• D. ${\tan }^{-1}\left(\frac{x^2+1}{x}\right)+C$

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+C$

$I=\int \frac{x^2+1}{x^4+x^2+1}dx$
Dividing $N^r$ and $D^r$ by $x^2$
$=\int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}dx$
$=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+3}dx$
Let $x-\frac{1}{x}=t$
$\Rightarrow (1+\frac{1}{x^2})dx=dt$
$\Rightarrow I=\int \frac{dt}{t^2+(\sqrt{3}{)}^2}\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}\Rightarrow I=\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{t}{\sqrt{3}}+C=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+C$