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JEE Main 4-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 4-April-2025 Paper - Shift 24 Marks
Question
$
\begin{array}{l}
\text { If } \int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} d x= \\
\frac{1}{m}\left(\left(\sqrt{1+x^2}+x\right)^n\left(n \sqrt{1+x^2}-x\right)\right)+C \text { where } C
\end{array}
$
is the constant of integration and $m , n \in N$, then $m + n$ is equal to
✓

Answer

379
rationalise
$\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 $
$\Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{1} d x$
Put $\sqrt{1+ x ^2}+ x = t$
$\left(\frac{x}{\sqrt{1+x^2}}+1\right) dx=dt$
$dx=\frac{dt}{t} \sqrt{1+x^2}$
Now as $\sqrt{1+ x ^2}+ x = t$
$\text { so } \sqrt{1+x^2}-x=\frac{1}{t} $
$\therefore \sqrt{1+x^2}=\frac{1}{2}\left(t+\frac{1}{t}\right)$
Thus $I=\int t^{19} \cdot \frac{ dt }{ t } \cdot \frac{1}{2}\left( t +\frac{1}{ t }\right)$
$\Rightarrow \frac{1}{2} \int\left(t^{19}+t^{17}\right) dt $
$=\frac{1}{2}\left(\frac{t^{20}}{20}+\frac{t^{18}}{18}\right)+C $
$=\frac{t^{19}}{360}\left[9 t+\frac{10}{t}\right]+C $
$=\frac{t^{19}}{360}\left[9\left(t+\frac{1}{t}\right)+\frac{1}{t}\right]+C $
$\Rightarrow \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{360}\left[9\left(2 \sqrt{1+x^2}\right)+\left(\sqrt{1+x^2}-x\right)\right]+C $
$\Rightarrow \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{360}\left[19 \sqrt{1+x^2}-x\right]+C $
$\therefore m=360, n=19 $
$\therefore m+n=379$

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