Question

$\int \frac{ dx }{ x ( x +1)(\ln ( x +1)-\ln x )^{11}}$ equals
where $C$ is constant of integration.

• A. $ \frac{1}{10(\ln ( x +1)-\ln x )^{10}}+ C$
• B. $\frac{(\ln ( x +1)-\ln x )^{10}}{10}+ C$
• C. $\frac{1}{11(\ln (x+1)-\ln x)^{11}}+C$
• D. $\frac{(\ln (x+1)-\ln x)^{11}}{11}+C$

Answer 1

Answer: (A)

Let $I=\int \frac{d x}{x(1+x)(\ln (x+1)-\ln x)^{11}}$
Put $\ln ( x +1)-\ln x = t \Rightarrow \frac{ dx }{ x (1+ x )}=- dt$
So, $I=-\int \frac{ dt }{ t ^{11}}=\frac{1}{10}\left(\frac{1}{ t ^{10}}\right)=\frac{1}{10(\ln ( x +1)-\ln x )^{10}}+ C$