Question

The value of $\int \frac{\ln (1+\frac{1}{x})}{x(x+1)}fx$ is

• A. $-\frac{1}{2}{\left(\ln (1+\frac{1}{x})\right)}^2+C$
• B. $\frac{1}{2}{\left(\ln (1+\frac{1}{x})\right)}^2+C$
• C. $-\frac{1}{2}{\left(\ln (1-\frac{1}{x})\right)}^2+C$
• D. $-\frac{1}{3}{\left(\ln (1+\frac{1}{x})\right)}^2+C$

Answer 1

The correct option is A $-\frac{1}{2}{\left(\ln (1+\frac{1}{x})\right)}^2+C$

Given,

$\int \frac{\ln (1+\frac{1}{x})}{x(x+1)}dx$

apply u substitution,

$u=\frac{1}{x}$

$=\int -\frac{\ln (u+1)}{1+u}du$

$=-\int \ln (u+1)\frac{1}{1+u}du$

apply v substitution

$v=\ln (1+u)$

$=-\int vdv$

$=-\frac{v^{1+1}}{1+1}$

substitute back

$v=\ln (1+u),u=\frac{1}{x}$

$=-\frac{{\ln }^{1+1}(1+\frac{1}{x})}{1+1}$

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

$=-\frac{1}{2}{\left(\ln (1+\frac{1}{x})\right)}^2+C$