JEE Main 2025 — Mathematics Calculus

rationalise

$\begin{aligned}
& \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} \times \frac{\left(\sqrt{1+x^2}+x\right)^9}{\left(\sqrt{1+x^2}+x\right)^9} d x \
& \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{1} d x
\end{aligned}$

Put $\sqrt{1+\mathrm{x}^2}+\mathrm{x}=\mathrm{t}$1+x2​+x=t

$$\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}+1\right) \mathrm{dx}=\mathrm{dt}$$(1+x2​x​+1)dx=dt

$$\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{t}} \sqrt{1+\mathrm{x}^2}$$dx=tdt​1+x2​

Now as $\sqrt{1+\mathrm{x}^2}+\mathrm{x}=\mathrm{t}$1+x2​+x=t

$\begin{aligned}
& \text { so } \sqrt{1+\mathrm{x}^2}-\mathrm{x}=\frac{1}{\mathrm{t}} \
& \therefore \sqrt{1+\mathrm{x}^2}=\frac{1}{2}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)
\end{aligned}$

Thus $\displaystyle I=\int \mathrm{t}^{19} \cdot \frac{\mathrm{dt}}{\mathrm{t}} \cdot \frac{1}{2}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)$I=∫t19⋅tdt​⋅21​(t+t1​)

$\begin{aligned}
& \Rightarrow \frac{1}{2} \int\left(\mathrm{t}^{19}+\mathrm{t}^{17}\right) \mathrm{dt} \
& =\frac{1}{2}\left(\frac{\mathrm{t}^{20}}{20}+\frac{\mathrm{t}^{18}}{18}\right)+\mathrm{C} \
& =\frac{\mathrm{t}^{19}}{360}\left[9 \mathrm{t}+\frac{10}{\mathrm{t}}\right]+\mathrm{C} \
& =\frac{\mathrm{t}^{19}}{360}\left[9\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)+\frac{1}{\mathrm{t}}\right]+\mathrm{C} \
& \Rightarrow \frac{\left(\sqrt{1+\mathrm{x}^2}+\mathrm{x}\right)^{19}}{360}\left[9\left(2 \sqrt{1+\mathrm{x}^2}\right)+\left(\sqrt{1+\mathrm{x}^2}-\mathrm{x}\right)\right]+\mathrm{C} \
& \Rightarrow \frac{\left(\sqrt{1+\mathrm{x}^2}+\mathrm{x}\right)^{19}}{360}\left[19 \sqrt{1+\mathrm{x}^2}-\mathrm{x}\right]+\mathrm{C} \
& \therefore \mathrm{m}=360, \mathrm{n}=19 \
& \therefore \mathrm{m}+\mathrm{n}=379
\end{aligned}$