Q. The value of the integral $\int_{}^{}\frac{dx}{x\left( 1 + \log x \right)^{2}}$ is equal to
A) $\frac{- 1}{1 + x} + c$
B) $\frac{- 1}{1 + \log x} + c$
C) $\frac{1}{1 + \log x} + c$
D) $\frac{1}{1 + x} + c$
Explanation
(b)
Put $1 + \log x = t\ \Rightarrow \ \frac{1}{x}dx = dt$
$\int_{}^{}{\frac{dx}{x\left( 1 + \log x \right)^{2}} = \int_{}^{}t^{- 2}dt = \frac{t^{- 1}}{- 1} + c = \frac{- 1}{(1 + \log x)} + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.