Question

$\int \frac{x^2-1}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}dx$ is equal to

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

Answer 1

The correct option is D $\log \left[{\tan }^{-1}(x+\frac{1}{x})\right]+C$

Let, $I=\int \frac{x^2-1}{(x^4+3x^2+1){\tan }^{-1}(x+\frac{1}{x})}dx$
On dividing numerator and denominator by $x^2$, we get
$I=\int \frac{1-\frac{1}{x^2}}{(x^2+\frac{1}{x^2}+3){\tan }^{-1}(x+\frac{1}{x})}dx$
Put $x+\frac{1}{x}=t$
$\Rightarrow (1-\frac{1}{x^2})dx=dt$ and
$\Rightarrow x^2+\frac{1}{x^2}=t^2-2$
Therefore, $I=\int \frac{dt}{(t^2-2+3){\tan }^{-1}t}$
$=\int \frac{dt}{(1+t{)}^2{\tan }^{-1}t}$
$=\log \left({\tan }^{-1}t\right)+C$
$=\log \left[{\tan }^{-1}(x+\frac{1}{x})\right]+C$.