Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
$\int \frac{(x + \sqrt{1 + x^2})^{15}}{\sqrt{1 + x^2}} dx$ is equal to:
\frac{(x + \sqrt{1 + x^2})^{16}}{10} + c
\frac{1}{15(\sqrt{1 + x^2} + x)} + c
\frac{15}{(\sqrt{1 + x^2} - x)} + c
\frac{(\sqrt{1 + x^2} + x)^{15}}{15} + c
Step-by-Step Solution
Key Concept: Substitution using the given relationship between $u$, $x$, and $s$ simplifies the integral to a power function.
Using the given notation $\sqrt{1+x^2} = (u-x)ds$ with $m = 1 > 0$ and $s = 4 > 0$, we substitute to get $I = \int \frac{u^{15}}{u}du = \int u^{14}du = \frac{1}{15}u^{15} + c$.
Correct Answer: 4