\(\int \frac{e^{e^mtan^{-1}x}}{(1 + x^2)^{3/2}} dx\)
Let \(tan^{-1}x = t\)
\(\frac{dx}{1 + x^2} = dt\)
Also
\(x = tan \,t\)
\(\therefore 1 + x^2= 1 + tan^2 t = sec^2 t\)
\(\therefore \int \frac{e^{e^mtan^{-1}x}}{(1 + x^2)^{3/2}} dx = \int\frac{e^{mt}dt}{sec\,t} = \int e^{mt} cos\,t\;dt = I\) (Let)
\(I = cos \,t\frac{e^{mt}}m + \int sin\,t \frac{e^{mt}}ndt\)
\(= cos\,t\frac{e^{mt}}m + sin\,t \frac{e^{mt}}{m^2} - \int \frac{cos\,t\,e^{mt}}{m^2}dt\)
\(I + \frac1{m^2}I= \frac{(mcos\,t + sin\,t)e^{mt}}{m^2}\)
⇒ \(I = \frac{(mcos\,t + sin\,t)e^{mt}}{1+m^2}\)
\(\therefore\int \frac{e^{e^mtan^{-1}x}}{(1 + x^2)^{3/2}} dx = \frac{mcos(tan^{-1}x)+ sin(tan^{-1}x)}{1 + m^2} e^{mtan^{-1}x}\)
\(=\frac{\frac m{\sqrt {1 +x^2}} + \frac x{\sqrt{1 + x^2}}}{1 + m^2}e^{mtan^{-1}x}\)
\(= \frac{m + x}{(1 + m^2)\sqrt{1 + x^2}} e^{mtan^{-1}x}\)
