Question

Evaluate: $\int x\log (1+x)dx$

• A. $\frac{(2x^2+2)\ln (x+1)-x^2+2x}{4}+c$
• B. $\frac{(2x^2+2)\ln (x+1)-x^2-2x}{2}+c$
• C. $\frac{(2x^2-2)\ln (x+1)-x^2+2x}{4}+c$
• D. $\frac{(2x^2-2)\ln (x+1)-x^2+4x}{2}+c$

Answer 1

Answer: (C)

$\frac{(2x^2-2)\ln (x+1)-x^2+2x}{4}+c$

$\int x\log (1+x)dx$

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

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

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

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

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

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

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

Hence, the answer is $\frac{(2x^2-2)\ln (x+1)-x^2+2x}{4}+c$