Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
MATCH THE FOLLOWING:
(A) $\int \frac{e^{2x} - 1}{e^{2x} + 1} dx$ is equal to
(B) $\int \frac{1}{(e^x + e^{-x})^2} dx$ is equal to
(C) $\int \frac{e^x}{1 + e^x} dx$ is equal to
(D) $\int \frac{1}{\sqrt{1 - e^{2x}}} dx$ is equal to
x - ln(e^{2x} + 1) + C
\frac{1}{2}\tanh(x) + C
ln(e^x + 1) + C
\sin^{-1}(e^x) + C
Step-by-Step Solution
Key Concept: Decomposing the integrand into a simple polynomial plus a rational function reduces the problem to recognizing standard forms.
For part (A), rewrite $\frac{e^{2x}-1}{e^{2x}+1}$ by separating numerator as $\frac{e^{2x}+1-2}{e^{2x}+1} = 1 - \frac{2}{e^{2x}+1}$. This gives $\int dx - \int \frac{2dx}{e^{2x}+1} = x - \int \frac{e^{-x}-e^x}{e^{-x}+e^x}dx = x - \ln(e^x + e^{-x}) + C = \ln(e^x + e^{-x}) + C$.
Correct Answer: 4