Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
If $I_n = \int \frac{dx}{(x^2 + a^2)^n}$, where $n \in \mathbb{N}$ and $n > 1$. If in and $I_{n-1}$ are related by the relation $PI_n = \frac{x}{(x^2 + a^2)^{n-1}} + QI_{n-1}$. Then $P$ and $Q$ are respectively given by:
$(2n - 1)a^2, 2n - 3
$2a^2(n - 1), 2n - 1
$a^2(n + 1), 2n + 3
$a^2, a^2(n + 1)$
Step-by-Step Solution
Key Concept: Integration by parts combined with algebraic manipulation produces a recurrence relation between consecutive reduction formula terms.
For $I_n = \int \frac{dx}{(x^2+a^2)^n}$, integrate by parts with $u$ as the second function. This yields $I_n = \frac{x}{(x^2+a^2)^n} + 2n\int \frac{x^2+a^2-a^2}{(x^2+a^2)^{n+1}}dx = \frac{x}{(x^2+a^2)^n} + 2n(I_n - a^2I_{n+1})$.
Correct Answer: 2