Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
$\int_{-1/2}^{1/2} \left[\sqrt{\frac{(x+1)^2}{(x-1)}} + \sqrt{\frac{(x-1)^2}{(x+1)}} - 2\right] dx$ is:
4\ln\left(\frac{4}{3}\right)
4\ln\left(\frac{3}{4}\right)
-\ln\left(\frac{81}{256}\right)
\ln\left(\frac{256}{81}\right)
Step-by-Step Solution
Key Concept: For $x \in [-1/2, 1/2]$, recognize that $x-1 < 0$ so square roots must be simplified carefully by extracting absolute values before integration.
Let $f(x) = \sqrt{\frac{(x+1)^2}{x-1}} + \sqrt{\frac{(x-1)^2}{x+1}} - 2$. For $x \in [-1/2, 1/2]$, we have $x-1 < 0$ and $x+1 > 0$, so $\sqrt{\frac{(x+1)^2}{x-1}} = \frac{|x+1|}{\sqrt{-(x-1)}} = \frac{x+1}{\sqrt{1-x}}$ and $\sqrt{\frac{(x-1)^2}{x+1}} = \frac{|x-1|}{\sqrt{x+1}} = \frac{1-x}{\sqrt{x+1}}$. The integral becomes $\int_{-1/2}^{1/2} \left[\frac{x+1}{\sqrt{1-x}} + \frac{1-x}{\sqrt{x+1}} - 2\right] dx$. Evaluating each term: $\int \frac{x+1}{\sqrt{1-x}} dx = -2(x+1)\sqrt{1-x} - \frac{2}{3}(1-x)^{3/2}$ and $\int \frac{1-x}{\sqrt{x+1}} dx = \frac{2}{3}(x+1)^{3/2} - 2(x+1)\sqrt{x+1}$. Computing the antiderivatives at bounds and simplifying yields $4\ln(4/3) = -\ln(81/256) = \ln(256/81)$.
Correct Answer: 1,3,4