If ∫ ( 1+x^2 +x)^ 10 ( 1+x^2 -x)^9 d x= (1)/(m)(( 1+x^2 +x)^n(n...

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