$\int e^{x}\left(\tan^{-1}x+\dfrac{1}{1+x^2}\right)dx$ equals
A$e^{x}\tan^{-1}x+C$
B$e^{x}\left(\tan^{-1}x-\dfrac{1}{1+x^2}\right)+C$
C$\tan^{-1}x+C$
D$\dfrac{e^{x}}{1+x^2}+C$
Answer & Solution
Correct answer: A. $e^{x}\tan^{-1}x+C$
1. Let $f(x)=\tan^{-1}x$, then $f'(x)=\dfrac{1}{1+x^2}$.
2. The integrand is $e^{x}[f(x)+f'(x)]$.
3. Using $\int e^{x}[f(x)+f'(x)]\,dx=e^{x}f(x)+C$.
4. Result: $e^{x}\tan^{-1}x+C$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.328_
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