Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
$I = \int \frac{(x^2-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)^2}dx$ is equal to (where $t = x + \frac{1}{x}$)
\sqrt{t^2+2t-3} - \ln\left|t+1+\sqrt{t^2+2t-3}\right| - \sqrt{3}\sin^{-1}\left(\frac{t+5}{2t+4}\right) + c
\sqrt{t^2+t-3} - \ln\left|t+1+\sqrt{t^2+t-3}\right| - \sqrt{3}\sin^{-1}\left(\frac{t+5}{2t+4}\right) + c
\sqrt{t^2+2t-3} - \ln\left|t+1+\sqrt{t^2+2t-3}\right| + \sqrt{3}\cos^{-1}\left(\frac{t+5}{2t+4}\right) + c
None of these
Step-by-Step Solution
Key Concept: The substitution $t = x + \frac{1}{x}$ reduces the degree of the denominator and simplifies the radical expression significantly.
The integral $I = \int \frac{(x^2-1)\sqrt{x^4+2x^2-x^2+2x+1}}{x^2(x+1)^2} dx$ is simplified by factoring the radicand and dividing numerator by denominator. Set $t = x + \frac{1}{x}$ so $dt = \left(1 - \frac{1}{x^2}\right) dx$. The integral transforms to $I = \int \frac{(t^2-2)+2t-1}{(t+2)} dt = \int \frac{\sqrt{t^2+2t-3}}{(t+2)} dt$. Through substitution and partial fraction decomposition, this yields $I = \sqrt{t^2+2t-3} - 3\int \frac{dt}{(t+2)\sqrt{t^2+2t}}$, which evaluates to the final antiderivative in terms of $t$.
Correct Answer: 1,3