Question

Solve: $\int \frac{x^2-1}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}dx$

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

Answer 1

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

$I=\int \frac{x^2-1}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}dx$

$=\int \frac{x^2(1-\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2}+3){\tan }^{-1}(x+\frac{1}{x})}dx$

$=\int \frac{1-\frac{1}{x^2}}{(x^2+\frac{1}{x^2}+3){\tan }^{-1}(x+\frac{1}{x})}dx$

Substitute $x+\frac{1}{x}=t$ or $(1-\frac{1}{x^2})dx=dt$ and $x^2+\frac{1}{x^2}+2=t^2$

$\therefore I=\int \frac{dt}{(t^2+1){\tan }^{-1}t}=\ln \left|{\tan }^{-1}t\right|+C$ $[\because \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log |f\left(x\right)|+C]$

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