Q. $\int_{}^{}{\frac{1 + x}{x + e^{- x}}\ dx}$ is equal to
A) $\log{\left| \left( x - e^{- x} \right) \right| + c}$
B) $\log{\left| \left( x + e^{- x} \right) \right| + c}$
C) $\log{\left| \left( 1 + xe^{x} \right) \right| + c}$
D) $\left( 1 + xe^{x} \right)^{2} + c$
Explanation
(c)
Let $I = \int_{}^{}{\frac{1 + x}{x + e^{- x}}\ dx} = \int_{}^{}{\frac{e^{x}(x + 1)}{{xe}^{x} + 1}\ dx}$
Put $xe^{x} + 1 = t$
$\Rightarrow \left( xe^{x} + e^{x} \right)dx = dt$
$\Rightarrow (x + 1)e^{x}\ dx = dt$
$\therefore\ I = \int_{}^{}\frac{dt}{t} = \log t + c$
$= \log{\left| \left( xe^{x} + 1 \right) \right| + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.