Question

$\int \frac{e^{m{\tan }^{-1}x}}{{\left(1+x^2\right)}^{3/2}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{e^{m{\tan }^{-1}x}}{{\left(1+x^2\right)}^{{\displaystyle \frac{3}{2}}}}dx \\ \mathrm{Putting}{\tan }^{-1}x=t\Rightarrow x=\tan t \\ \Rightarrow \frac{1}{1+x^2}dx=dt \\ \Rightarrow dx=\left(1+x^2\right)dt \\ \Rightarrow dx=\left(1+{\tan }^{2}t\right)dt \\ \therefore I=\int \frac{e^{mt}}{{\left(1+{\tan }^{2}t\right)}^{{\displaystyle \frac{3}{2}}}}\left(1+{\tan }^{2}t\right)dt \\ =\int \frac{e^{mt}dt}{\sqrt{1+{\tan }^{2}t}} \\ =\int {\underset{\mathrm{II}}{e}}^{mt}\underset{\mathrm{I}}{\cos }tdt \\ =\cos t\frac{e^{mt}}{m}-\int \left(-\sin t\right)\frac{e^{mt}}{m}dt \\ =\cos t\frac{e^{mt}}{m}+\frac{1}{m}\int e^{mt}\sin tdt \\ =\cos t\frac{e^{mt}}{m}+\frac{1}{m}{\mathrm{I}}_1.....\left(1\right) \\ \mathrm{Where}, \\ {I}_1=\int {\underset{\mathrm{II}}{e}}^{mt}\underset{\mathrm{I}}{\sin t}dt \\ =\sin t\frac{e^{mt}}{m}-\int \cos t\frac{e^{mt}}{m}dt \\ {I}_1=\sin t\frac{e^{mt}}{m}-\frac{1}{m}I\mathit{}\mathit{}.....\left(2\right) \\ \mathrm{from}\left(1\right)\mathrm{and}\left(2\right) \\ I=\cos t\frac{e^{mt}}{m}+\frac{1}{m}\left[\sin t\frac{e^{mt}}{m}-\frac{1}{m}I\right] \\ \Rightarrow I=\cos t\frac{e^{mt}}{m}+\frac{\sin t{e}^{mt}}{m^2}-\frac{1}{m^2}\mathrm{I} \\ \Rightarrow I+\frac{I}{m^2}=\frac{e^{mt}\left(m\cos t+\sin t\right)}{m^2} \\ \Rightarrow I=\frac{e^{mt}\left(m\cos t+\sin t\right)}{1+m^2}+C \\ \Rightarrow I=\frac{e^{mt}}{\sqrt{1+m^2}}\left[\cos t\frac{m}{\sqrt{1+m^2}}+\sin t\frac{1}{\sqrt{1+m^2}}\right]+C \\ \mathrm{Let}\frac{m}{\sqrt{1+m^2}}=\cos \theta \\ \mathrm{Then},\sin \theta =\frac{1}{\sqrt{1+m^2}} \\ \Rightarrow \cot \theta =m \\ \Rightarrow \theta ={\cot }^{-1}m \\ \therefore I=\frac{e^{mt}}{\sqrt{1+m^2}}\left\{\cos t\cos \theta +\sin t\sin \theta \right\}+C \\ =\frac{e^{mt}}{\sqrt{1+m^2}}\left\{\cos \left(t-\theta \right)\right\}+C \\ =\frac{e^{mt}}{\sqrt{1+m^2}}\left\{\cos \left({\tan }^{-1}x-{\cot }^{-1}m\right)\right\}+C \end{gathered}$$