Question

$\int \frac{x^2-1}{x^4+3x^2+1}dx\left(x>0\right)$ is

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

Answer 1

Answer: (A)

Let $I=\int \frac{x^2-1}{x^4+3x^2+1}dx$
$=\int \frac{1-1/x^2}{x^2+3+1/x^2}dx$
$=\int \frac{1-1/x^2}{\left(x^2+\frac{1}{x^2}\right)+3}dx$
$=\int \frac{1-1/x^2}{{\left(x+\frac{1}{x}\right)}^2-2+3}dx$
$=\int \frac{1-1/x^2}{{\left(x+\frac{1}{x}\right)}^2+1}dx$
Let $x+\frac{1}{x}=t$
$\Rightarrow \left(1-\frac{1}{x^2}\right)dx=dt$
$\therefore I=\int \frac{dt}{t^2+1}$
$={\tan }^{-1}t+C$
$={\tan }^{-1}\left(x+\frac{1}{x}\right)+C\left[\because t=x+\frac{1}{x}\right]$