Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
$$\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} dx \text{ for } x > 0 \text{ is equal to:}$$
$$2\tan^{-1}\sqrt{x^2 + \frac{1}{x^2} + 1} + c$$
$$\tan^{-1}\sqrt{x + \frac{1}{x} + 1} + c$$
$$2\tan^{-1}\sqrt{x + \frac{1}{x} + 1} + c$$
$$2\sec^{-1}\sqrt{x + \frac{1}{x} + 2} + c$$
Step-by-Step Solution
Key Concept: Recognizing that $d(\sqrt{x + \frac{1}{x} + 1}) \propto \frac{x-1}{\sqrt{x}\sqrt{x^2+x+1}}$ allows direct substitution leading to a standard arctangent form.
First, factor the expression under the square root: $x^3 + x^2 + x = x(x^2 + x + 1)$, so $\sqrt{x^3 + x^2 + x} = \sqrt{x}\sqrt{x^2 + x + 1}$. The integral becomes $\int \frac{x-1}{(x+1)\sqrt{x}\sqrt{x^2 + x + 1}} dx$. Use substitution $u = \sqrt{x + \frac{1}{x} + 1}$, which gives $du = \frac{1}{2}\frac{x-1}{\sqrt{x}\sqrt{x^2+x+1}} dx$. This transforms the integral to $\int \frac{2du}{u^2+1} = 2\tan^{-1}(u) + c = 2\tan^{-1}\sqrt{x + \frac{1}{x} + 1} + c$. The fourth option is equivalent since $\tan^{-1}(u) = \sec^{-1}(\sqrt{u^2+1})$ for appropriate ranges, giving $2\sec^{-1}\sqrt{x + \frac{1}{x} + 2} + c$.
Correct Answer: 3,4