Question

$\displaystyle \int \frac{x^{2} - 2}{x^{4} + 4}dx$ is equal to (where $C$ is constant of integration)

• A. $\frac{1}{4}log\left|\frac{x^{2} + 2 x + 2}{x^{2} - 2 x + 2}\right|+C$
• B. $\frac{1}{4}log\left|\frac{x^{2} - 2 x + 2}{x^{2} + 2 x + 2}\right|+C$
• C. $\frac{1}{2}\left(tan\right)^{- 1}\left(\frac{x^{2} + 2 x + 2}{x^{2} - 2 x + 2}\right)+C$
• D. $\frac{1}{2}\left(tan\right)^{- 1}\left(\frac{x^{2} - 2 x + 2}{x^{2} + 2 x + 2}\right)+C$

Answer 1

Answer: (B)

$\displaystyle \int \frac{x^{2} - 2}{x^{4} + 4}dx\Rightarrow \displaystyle \int \frac{1 - \frac{2}{x^{2}}}{x^{2} + \frac{4}{x^{2}}}dx$
$\displaystyle \int \frac{1 - \frac{2}{x^{2}}}{\left(x + \frac{2}{x}\right)^{2} - 4}dx$
Put $\left(x + \frac{2}{x}\right)=t$
$\displaystyle \int \frac{d t}{t^{2} - 4}$
$\frac{1}{4}log\left|\frac{t - 2}{t + 2}\right|+C=\frac{1}{4}log\left|\frac{x^{2} - 2 x + 2}{x^{2} + 2 x + 2}\right|+C$