Question

For \(x\neq 0\), the value of \(I=\int \frac{x^{2}-1}{(x^{4}+3x^{2}+1)\tan ^{-1}\left(x+\frac{1}{x}\right)}dx\) is ...

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

Hint:

Divide numerator and denominator by x squared and substitute u = x + 1/x.

Answer 1

The Correct Option is A

Step 1: Differentiate to verify
Take $F=\log \left(\tan ^{-1}u\right)$ with $u=x+\dfrac{1}{x}$.

Step 2: Chain rule
$F'=\dfrac{1}{\tan ^{-1}u}\cdot \dfrac{1}{1+u^{2}}\cdot \left(1-\dfrac{1}{x^{2}}\right)$.

Step 3: Simplify
$1+u^{2}=x^{2}+3+\dfrac{1}{x^{2}}=\dfrac{x^{4}+3x^{2}+1}{x^{2}}$, so $F'=\dfrac{x^{2}-1}{(x^{4}+3x^{2}+1)\tan ^{-1}u}$, the given integrand.

Final Answer:
Differentiating the first option returns the integrand. \[ \boxed{\text{(A) }\log \left(\tan ^{-1}\left(x+\dfrac{1}{x}\right)\right)+c} \]