Question

The value of $\int \frac{\ln n\left(1-\left(\frac{1}{x}\right)\right)dx}{x\left(x-1\right)}$ is

• A. $\frac{1}{2}\left[m{\left(1-\frac{1}{x}\right)}^2\right]+C$
• B. $\frac{1}{2}{\left[m\left(1+\frac{1}{x}\right)\right]}^2+C$
• C. $\frac{1}{2}\ell n\left(x\left(x-1\right)\right)+C$
• D. $\frac{1}{2}{\left[\ell n\left(x\right(x-1\left)\right)\right]}^2+C$

Answer 1

Answer: (A)

$\frac{1}{2}\left[m{\left(1-\frac{1}{x}\right)}^2\right]+C$

Consider, $I=\int \frac{ln\left(1-\frac{1}{x}\right)}{x\left(x-1\right)}dx$

$I=\int \frac{\ln \left(1-\frac{1}{x}\right)}{x^2\left(1-\frac{1}{x}\right)}dx$

Put $\ln \left(1-\frac{1}{x}\right)=t$ $\Rightarrow$ $\frac{1}{\left(1-\frac{1}{x}\right)}\times \frac{1}{x^2}dx=dt$

$I=\int tdt$

$I=\frac{t^2}{2}+c$

$I=\frac{1}{2}{\left[ln\left(1-\frac{1}{x}\right)\right]}^2+c$