Q. $\int_{}^{}{\frac{(x + 3)e^{x}}{(x + 4)^{2}}\ dx}$ is equal to
A) $\frac{1}{(x + 4)^{2}} + c$
B) $\frac{e^{x}}{(x + 4)^{2}} + c$
C) $\frac{e^{x}}{x + 4} + c$
D) $\frac{e^{x}}{x + 3} + c$
Explanation
(c)
$\int_{}^{}{\frac{(x + 3)e^{x}}{(x + 4)^{2}}\ dx} = \int_{}^{}\frac{(x + 4 - 1)e^{x}dx}{(x + 4)^{2}}$
$= \int_{}^{}{\left( \frac{x + 4}{(x + 4)^{2}} - \frac{1}{(x + 4)^{2}} \right)e^{x}dx}$
$= \int_{}^{}{e^{x}\frac{1}{x + 4}dx - \int_{}^{}{e^{x}\frac{1}{(x + 4)^{2}}dx}}$
$= \frac{e^{x}}{x + 4} + \int_{}^{}{e^{x}\frac{1}{(x + 4)^{2}}dx - \int_{}^{}{\frac{e^{x}}{(x + 4)^{2}}dx + c}}$
$= \frac{e^{x}}{x + 4} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.