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