Q. If $\int_{}^{}{\frac{e^{x} - 1}{e^{x} + 1}\ dx = f(x) + c}$, then $f(x)$ is equal to
A) $2\log{(e^{x} + 1)}$
B) $\log{(e^{2x} - 1)}$
C) $2\log{\left( e^{x} + 1 \right) - x}$
D) $\log{(e^{2x} + 1)}$
Explanation
(c)
$\int_{}^{}{\frac{e^{x} - 1}{e^{x} + 1}\ dx = \int_{}^{}{\left( \frac{2e^{x}}{e^{x} + 1} - 1 \right)\ dx}}$
$= 2\log{\left( e^{x} + 1 \right) - x + c = f(x) + c}$ [given]
$\therefore\ f(x) = 2\log{(e^{x} + 1)} - x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.