Q. $\int_{}^{}{\frac{(x^{3} + 3x^{2} + 3x + 1)}{(x + 1)^{5}}\ dx}$ is equal to
A) $- \frac{1}{(x + 1)} + c$
B) $\frac{1}{5}\log{(x + 1) + c}$
C) $\log{(x + 1) + c}$
D) $\tan^{- 1}{x + c}$
Explanation
(a)
Let $I = \int_{}^{}{\frac{(x^{3} + 3x^{2} + 3x + 1)}{(x + 1)^{5}}\ dx}$
$= \int_{}^{}\frac{(x + 1)^{3}}{(x + 1)^{5}}dx = \int_{}^{}\frac{1}{(x + 1)^{2}}\ dx$
$= - \frac{1}{(x + 1)} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.