Question

The value of $\int \frac{x^2+1}{x^4-x^2+1}dx$ is

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

Answer 1

Answer: (E)

Let $I=\int \frac{x^2+1}{x^4-x^2+1}dx$
$=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-1}dx$
$=\int \frac{\left(1+\frac{1}{x^2}\right)}{{\left(x-\frac{1}{x}\right)}^2+1}dx$ Let $x-\frac{1}{x}=t$ and $\left(1+\frac{1}{x^2}\right)dx=dt$
$\therefore$ $I=\int \frac{dt}{t^2+1}={\tan }^{-1}t+c$
$={\tan }^{-1}\left(x-\frac{1}{x}\right)+c$
$={\tan }^{-1}\left(\frac{x^2-1}{x}\right)+c$