Question

If $\int x\ln (1+\frac{1}{x})dx=f\left(x\right)\ln (1+\frac{1}{x})+A\ln |x+1|+Bx+C$, then which of the following is(are) correct
(where $C$ is constant of integration)

• A. $f\left(x\right)=\frac{x^2}{2}$
• B. $f\left(x\right)=\frac{x^2}{2}-\frac{1}{2}$
• C. $\frac{A}{B}=-1$
• D. $\frac{A}{B}=-4$

Answer 1

Answer: (A), (C)

$\frac{A}{B}=-1$

$\int x\cdot \ln (1+\frac{1}{x})dx\underset{II}{\downarrow }\underset{I}{\downarrow }$
Applying integration by parts
$=\frac{x^2}{2}\ln (1+\frac{1}{x})-\int \frac{x}{1+x}\cdot (-\frac{1}{x^2})\cdot \frac{x^2}{2}dx=\frac{x^2}{2}\ln (1+\frac{1}{x})+\frac{1}{2}\int \frac{x}{1+x}dx=\frac{x^2}{2}\ln (1+\frac{1}{x})+\frac{1}{2}\int \frac{x+1-1}{1+x}dx=\frac{x^2}{2}\ln (1+\frac{1}{x})+\frac{1}{2}\int [1-\frac{1}{1+x}]dx=\frac{x^2}{2}\ln (1+\frac{1}{x})-\frac{1}{2}\ln |1+x|+\frac{x}{2}+C$
Hence, $f\left(x\right)=\frac{x^2}{2},A=-\frac{1}{2}$ and $B=\frac{1}{2}$