Question

$\int \frac{(x^2-1)}{(x^4+3x^2+1){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}dx=\ln \left|f\right(x\left)\right|+C$ then f(x) is

• A. $\ln |x+\frac{1}{x}|$
• B. ${\tan }^{-1}|x+\frac{1}{x}|$
• C. ${\cot }^{-1}|x+\frac{1}{x}|$
• D. $\ln \left|{\tan }^{-1}(x+\frac{1}{x})\right|$

Answer 1

Answer: (B)

${\tan }^{-1}|x+\frac{1}{x}|$

$I=\int \left(\frac{x^2-1}{(x^4+3x^2+1)\left({\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)}\right)dx$
Substitute ${\tan }^{-1}\left(\frac{x^2+1}{x}\right)=t$ and $\frac{x^2-1}{(x^4+3x^2+1)}dx=dt$
Therefore,
$I=\int \frac{dt}{t}=\ln \left|t\right|+C=\ln \left|{\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right|+C$

$=\ln \left|{\tan }^{-1}(x+\frac{1}{x})\right|+C$