JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 21)
If $\int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} \mathrm{~d} x=\frac{1}{\mathrm{~m}}\left(\left(\sqrt{1+x^2}+x\right)^{\mathrm{n}}\left(\mathrm{n} \sqrt{1+x^2}-x\right)\right)+\mathrm{C}$ where C is the constant of integration and $\mathrm{m}, \mathrm{n} \in \mathbf{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to _________ .
Answer
379
Explanation
$$\begin{aligned} & \sqrt{1+x^2}+x=\sec \theta+\tan \theta=t \\ & \sqrt{1+x^2}=\sec \theta=\frac{t^2+1}{2 t} \\ & x=\tan \theta=\frac{t^2-1}{2 t} \end{aligned}$$
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
$$\begin{aligned} & n=19 \\ & m=360 \\ & \therefore \quad n+m=379 \end{aligned}$$
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