Question

The value of $\int \frac{e^{x}\left(x^{2} \tan ^{-1} x + \tan^{-1} x+1\right)}{x^{2} + 1} dx $ is equal to

• A. $e^x \tan^{-1}x + c$
• B. $-\tan^{-1}x(e^x) + c$
• C. $\tan^{-1}(x^e) + c$
• D. $e^{\tan^{-1}x} + c$

Answer 1

Answer: (A)

Let $I=\int e^{x}\left(\frac{x^{2} \tan ^{-1} x+\tan ^{-1} x+1}{x^{2}+1}\right) d x$
$\Rightarrow I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{x^{2}+1}\right) d x$
If $f(x)=\tan ^{-1} x$, then
$f'(x)=\frac{1}{x^{2}+1}$
$\therefore I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{x^{2}+1}\right) d x$
$=e^{x} \tan ^{-1} x+C$
$\left[\because \int e^{x}\left[f(x)+f'(x)\right] d x=e^{x} f(x)+C\right]$