allery$\int \dfrac{e^{\tan^{-1}x}}{1+x^2}\,dx$ equals
A$e^{x}\tan^{-1}x+C$
B$\dfrac{e^{\tan^{-1}x}}{1+x^2}+C$
C$\tan^{-1}\!\left(e^{x}\right)+C$
D$e^{\tan^{-1}x}+C$
Answer & Solution
Correct answer: D. $e^{\tan^{-1}x}+C$
1. Put $t=\tan^{-1}x$, so $dt=\dfrac{dx}{1+x^2}$.
2. Integral $=\int e^{t}\,dt=e^{t}$.
3. Back-substitute: $e^{\tan^{-1}x}+C$.
4. Check: $\dfrac{d}{dx}e^{\tan^{-1}x}=e^{\tan^{-1}x}\cdot\dfrac{1}{1+x^2}$, matching the integrand.
_Source: syllabus Mathematics Ch 7 "Integrals", p.302_
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