If int frac (√(1+x 2) + x) 10 (√(1+x 2) - x) 9 dx = 1/m ( (√(1+x 2) + x) n (√(1+x 2) - x) n ) + C where C is the constant of integration and m , n in mathbb N , then m+n is equal to \ \ \ \ -.

Mathematics · Indefinite Integration

JEE Main 2025 — 4 April, Evening Shift — Question 40

If $\int \frac{(\sqrt{1+x^2} + x)^{10}}{(\sqrt{1+x^2} - x)^{9}} dx $$ = \frac{1}{m} \left( (\sqrt{1+x^2} + x)^n (\sqrt{1+x^2} - x)^n \right) + C$ where $C$ is the constant of integration and $m$ , $n \in \mathbb{N}$ , then $m+n$ is equal to $\_\_\_\_$ -.

Answer: 379

Numerical answer — enter this value.

Step-by-step solution

$\sqrt{1+x^{2}}+x=\sec \theta+\tan \theta=t$

$\sqrt{1+x^{2}}=\sec \theta=\frac{t^{2}+1}{2 t}$

$x=\tan \theta=\frac{t^{2}-1}{2 t}$

The given expression becomes

$\frac{1}{m} t^{2}\left(n \cdot \frac{t^{2}+1}{2 t}-\frac{t^{2}-1}{2 t}\right)=\frac{t^{n-1}}{2 m}\left((n-1) t^{2}+n+1\right)$

By compare

$n=19$

$m=360$

$\therefore \quad n+m=379$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration
Topic: Miscellaneous Types of Integrals