Question

If $\int x$ log$(1+\frac{1}{x})dx=$ $f\left(x\right)\log (x+1)+g\left(x\right).x^2+Lx+c$, then

• A. $f\left(x\right)=\frac{x^2}{2}-\frac{1}{2},g\left(x\right)=\frac{1}{2}\log x,L=1$
• B. $f\left(x\right)=\frac{x^2}{2}+\frac{1}{2},g\left(x\right)=-\frac{1}{2}\log x,L=\frac{1}{2}$
• C. $f\left(x\right)=\frac{x^2}{2}-\frac{1}{2},g\left(x\right)=-\frac{1}{2}\log x,L=\frac{1}{2}$
• D. $f\left(x\right)=\frac{x^2}{2}-\frac{1}{2},g\left(x\right)=\frac{1}{2}\log x,L=-\frac{1}{2}$

Answer 1

The correct option is C $f\left(x\right)=\frac{x^2}{2}-\frac{1}{2},g\left(x\right)=-\frac{1}{2}\log x,L=\frac{1}{2}$

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

$=\log (1+\frac{1}{x})\cdot \frac{x^2}{2}-\int \left(\frac{d}{dx}\log \right(1+\frac{1}{x})\int xdx)\cdot dx$

$=\log (1+\frac{1}{x})\cdot \frac{x^2}{2}-\int \frac{x}{\left(\frac{x+1}{x}\right)}\cdot \frac{-1}{x}xdx.$

$=\log (1+\frac{1}{x})\cdot \frac{x^2}{2}+\int \frac{x}{x+1}\cdot dx.$

$=\log (1+\frac{1}{x})\cdot \frac{x^2}{2}+x-\log (x+1).$

$=\left[\log \right(1+x\left)\right]\frac{x^2}{2}$