Q. $\int_{}^{}\frac{x^{e - 1} + e^{x - 1}}{x^{e} + e^{x}}\ dx$ is equal to
A) $\log{(x^{e} + e^{x})} + c$
B) $e\ \log{(x^{e}} + e^{x}) + c$
C) $\frac{1}{e}\log{(x^{e} + e^{x}}) + c$
D) None of these
Explanation
(c)
Put $x^{e} + e^{x} = t\ \ \ \Rightarrow \ e\left( x^{e - 1} + e^{x - 1} \right)dx = dt$
$\therefore\int_{}^{}\frac{x^{e - 1} + e^{x - 1}}{x^{e} + e^{x}}dx = \frac{1}{e}\int_{}^{}\frac{dt}{t} = \frac{1}{e}\log t + c = \frac{1}{e}\log{(x^{e} + e^{x})} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.