Question

The value of $\int \frac{(x^2-1)}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}dx$ is

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

Answer 1

The correct option is D $\ln \left|{\tan }^{-1}(x+\frac{1}{x})\right|+C$

$\int \frac{(1-\frac{1}{x^2})}{({(1+\frac{1}{x})}^2+1){\tan }^{-1}(x+\frac{1}{x})}dx$
Let $(x+\frac{1}{x})=t$
$\Rightarrow (1-\frac{1}{x^2})dx=dt$
$\int \frac{dt}{(t^2+1){\tan }^{-1}t}$

Let ${\tan }^{-1}t=v$
$\Rightarrow \frac{dt}{1+t^2}=dv$
$\int \frac{dv}{v}=\ln v+C$
$=\ln \left|{\tan }^{-1}(x+\frac{1}{x})\right|+C$