Question

$\int \frac{\log (1-x)}{x^2}dx$ equal to

• A. $\left[\right(1+\frac{1}{x}\left)\log \right(1-x)-\log x]+c$
• B. $(1-\frac{1}{x}\left)\log \right(1-x)-\log x+c$
• C. $(\frac{1}{x}-1\left)\log \right(1-x)+\log x+c$
• D. $(\frac{1}{x}-1\left)\log \right(1-x)-\log x+c$

Answer 1

Answer: (B)

$(1-\frac{1}{x}\left)\log \right(1-x)-\log x+c$

$\frac{\int log(1-x)}{x^2}dx$

$log(1-x)\int 1/x^{2}dx-\int [\frac{1}{(1-x)}\int \frac{1}{x^2}dx]dx$

$\frac{-log(1-x)}{x}+c-\int \frac{1}{(1-x)}-\frac{1}{x}dx$

$=\frac{-log(1-x)}{x}-\int [\frac{1}{x}+\frac{1}{1-x}]dx+c$

$=\frac{-log(1-x)}{x}-logx+log(1-x)+c$

$=log(1-x)[1-\frac{1}{x}]-logx+c$

$\int \frac{log(1-x)}{x^2}dx-log(1-x)[1-1/x]-logx+c$