Question

$\int \frac{(x^2+1)}{x^4+7x^2+1}dx=$

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

Answer 1

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

Let $I=\int \frac{(x^2+1)}{x^4+7x^2+1}dx=\int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}}dx=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x}{)}^2+9}dx$

Substitute $x-\frac{1}{x}=u\Rightarrow \frac{du}{dx}=1+\frac{1}{x^2}$

Ergo $I=\int \frac{du}{u^2+3^2}=\frac{1}{3}{\tan }^{-1}\frac{u}{3}+c=\frac{1}{3}{\tan }^{-1}\left(\frac{x^2-1}{3x}\right)+c$