Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
$$\int \frac{x^2 - 1}{\left(x^4 + 3x^2 + 1\right)\tan^{-1}\left(x + \frac{1}{x}\right)} dx =$$
$$\tan^{-1}\left(x + \frac{1}{x}\right) + C$$
$$\left(x + \frac{1}{x}\right)\tan^{-1}\left(x + \frac{1}{x}\right) + C$$
$$\ln\tan^{-1}\left(x + \frac{1}{x}\right) + C$$
$$\frac{1}{2}\ln\left|x + \frac{1}{x}\right| + C$$
Step-by-Step Solution
Key Concept: Strategic substitution $t = x + \frac{1}{x}$ simplifies the complex denominator into a recognizable form.
Rewrite the integral by separating the numerator as $x^2 - 1 = (x^2 + 3) - 4$ to split into two parts. Substitute $t = x + \frac{1}{x}$, which gives $dt = (1 - \frac{1}{x^2})dx$ and $t^2 = x^2 + 2 + \frac{1}{x^2}$. The integral transforms to $\int \frac{dt}{(t^2+1)\tan^{-1}t}$, which evaluates to $\ln|\tan^{-1}t| + C$ or $\ln|\tan^{-1}(x + \frac{1}{x})| + C$.
Correct Answer: 3