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

Question

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

Aicharya · Accepted Answer

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

If $\int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} d x=$…

Solution