Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $\int \frac{\sqrt{1+x^{2n}}\ln(1+x^{2n}) - 2n\ln x}{x^{2n+1}} dx = \frac{\alpha p}{\beta n}[1-3\ln p] + c$ where $p = \sqrt{1+\frac{1}{x^{2n}}}, \alpha,\beta \in \mathbb{N}$ then $\alpha + \beta = $
Step-by-Step Solution
Key Concept: Recognize binomial expansion in the sum and convert it to an integral using the limit definition of $e^x$.
The limit $\lim_{n \to \infty} \sum_{k=0}^{n} \frac{C_k}{n^k(k+3)}$ is evaluated by recognizing that $\sum_{k=0}^{n} C_k \left(\frac{x}{n}\right)^k = \left(1 + \frac{x}{n}\right)^n \to e^x$ as $n \to \infty$. The sum is converted to an integral: $\int_0^1 x^2 \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n dx = \int_0^1 x^2 e^x dx$. Using integration by parts twice yields $\left[x^2 e^x\right]_0^1 - \int_0^1 2xe^x dx = e - 2$.
Correct Answer: 0.72