Mathematics - Indefinite Integration Question with Solution | TestHub

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