Question

Evaluate the integral
${\int }_0^1(1+x).\log (1+x)dx$

• A. $\log 4-\frac{3}{4}$
• B. $\log 2+\frac{3}{4}$
• C. $\frac{1}{2}\log 2-\frac{3}{4}$
• D. $log4+\frac{3}{4}$

Answer 1

The correct option is A $\log 4-\frac{3}{4}$

${\int }_0^1(1+x)log(1+x)dx$

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

$={[log\frac{(1+x)(1+x{)}^2}{2}-\frac{1}{4}(1+x{)}^2]}_0^1$

$=[log\frac{\left(2\right)H}{2}-\frac{1}{4}(2{)}^2]-[0-\frac{1}{4}]$

$=2log\left(2\right)-\frac{3}{4}$

$=log4-\frac{3}{4}$