JEE Main 2025 Apr 4 Shift 2, Mathematics Q22: Methods of Integration

Rationalise the denominator by multiplying numerator and denominator by $(\sqrt{1+x^2}+x{)}^9$:

$(\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x)=1$, so the integrand becomes $(\sqrt{1+x^2}+x{)}^{19}$.

Substitute $t=\sqrt{1+x^2}+x$. Then $dt=\left(\frac{x}{\sqrt{1+x^2}}+1\right)dx=\frac{t}{\sqrt{1+x^2}}dx$, so $dx=\frac{\sqrt{1+x^2}}{t}dt$.

Also, $\sqrt{1+x^2}-x=\frac{1}{t}$, so $\sqrt{1+x^2}=\frac{1}{2}\left(t+\frac{1}{t}\right)$.

$I=\int t^{19}\cdot \frac{1}{2}\left(t+\frac{1}{t}\right)\cdot \frac{dt}{t}=\frac{1}{2}\int (t^{19}+t^{17})dt=\frac{1}{2}\left(\frac{t^{20}}{20}+\frac{t^{18}}{18}\right)+C$.

Factor $\frac{t^{19}}{360}$:

$I=\frac{t^{19}}{360}\left[9t+\frac{10}{t}\right]+C=\frac{t^{19}}{360}\left[9\left(t+\frac{1}{t}\right)+\frac{1}{t}\right]+C$.

Now $t+\frac{1}{t}=2\sqrt{1+x^2}$ and $\frac{1}{t}=\sqrt{1+x^2}-x$, so

$I=\frac{(\sqrt{1+x^2}+x{)}^{19}}{360}\left[18\sqrt{1+x^2}+\sqrt{1+x^2}-x\right]+C=\frac{(\sqrt{1+x^2}+x{)}^{19}}{360}\left[19\sqrt{1+x^2}-x\right]+C$.

Matching the given form: $m=360$, $n=19$, hence $m+n=379$.