$\int \dfrac{e^{ an^{-1}x}}{1+x^2}\,dx$ equals

Question

$\int \dfrac{e^{	an^{-1}x}}{1+x^2}\,dx$ equals

Accepted Answer

D. $e^{	an^{-1}x}+C$

1. Put $t=	an^{-1}x$, so $dt=\dfrac{dx}{1+x^2}$.
2. Integral $=\int e^{t}\,dt=e^{t}$.
3. Back-substitute: $e^{	an^{-1}x}+C$.
4. Check: $\dfrac{d}{dx}e^{	an^{-1}x}=e^{	an^{-1}x}\cdot\dfrac{1}{1+x^2}$, matching the integrand.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.302_

Answer

$e^{x}	an^{-1}x+C$

Answer

$\dfrac{e^{	an^{-1}x}}{1+x^2}+C$

Answer

$	an^{-1}\!\left(e^{x}ight)+C$

$\int \dfrac{e^{\tan^{-1}x}}{1+x^2}\,dx$ equals