Question

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

• A. $\frac {1}{2}x e^{\tan^{-1}x}+c$
• B. $x e^{\tan^{-1}x}+c$
• C. $e^{\tan^{-1}x}+c$
• D. $\frac {1}{2}e^{\tan^{-1}x}+c$

Answer 1

Answer: (B)

Let $I=\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x$
$=\int e^{\tan ^{-1} x} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x$
$=x e^{\tan ^{-1} x}-\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x+c$
$=x e^{\tan ^{-1} x}+c$