Question

$\int \frac{(x^2-2)dx}{(x^4+5x^2+4){\tan }^{-1}\left(\frac{x^2+2}{x}\right)}$ is

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

Answer 1

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

Substitute $\frac{x^2+2}{x}=y\Rightarrow dy=(1-\frac{2}{x^2})dx$
$\therefore \int \frac{(x^2-2)dx}{(x^4+5x^2+4){\tan }^{-1}\left(\frac{x^2+2}{x}\right)}=\int \frac{x^2(1-\frac{2}{x^2})dx}{({(x^2+2)}^2+x^2){\tan }^{-1}\left(\frac{x^2+2}{x}\right)}$
$=\int \frac{dy}{(y^2+1){\tan }^{-1}y}$

we know that $\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log |f\left(x\right)+C|$

$=\log \left|{\tan }^{-1}y\right|=\log \left|{\tan }^{-1}(x+\frac{2}{x})\right|+c$