Question

The value of $\int \frac{d_x}{x^4+5x^2+4}$ is.

• A. $\left(1/3\right){\tan }^{-1}x-\left(1/2\right){\tan }^{-1}\left(x/2\right)+c$
• B. 3 ${\tan }^{-1}\left(x/3\right)+2{\tan }^{-1}\left(x/2\right)+c$
• C. $\left(1/3\right){\tan }^{-1}x+\left(1/6\right){\tan }^{-1}\left(x/2\right)+c$
• D. $\left(1/3\right){\tan }^{-1}x-\left(1/6\right){\tan }^{-1}\left(x/2\right)+c$

Answer 1

The correct option is D $\left(1/3\right){\tan }^{-1}x-\left(1/6\right){\tan }^{-1}\left(x/2\right)+c$

$\int \frac{dx}{x^4+5x^2+4}=\int \frac{dx}{(x^2+4)(x^2+1)}$
$=\frac{1}{3}\int \frac{1}{(x^2+1)}-\frac{1}{(x^2+4)}dx$
$=\frac{1}{3}[\int \frac{1}{(x^2+1)}dx-\int \frac{1}{x^2+4}dx]$
$=\frac{1}{3}{\tan }^{-1}x-\frac{1}{6}{\tan }^{-1}\left(\frac{x}{2}\right)+c$