Question

${\int }_0^{1}xlog(1+2x)dx$

Answer 1

Let $I={\int }_0^{1}xlog(1+2x)dx$

$={\left[log\right(1+2x\left)\frac{x^2}{2}\right]}_0^1-\int \frac{1}{1+2x}.2.\frac{x^2}{2}dx=\frac{1}{2}\left[x^{2}log\right(1+2x){]}_0^1-\int \frac{x^2}{1+2x}dx=\frac{2}{2}[1log3-0]-\left[{\int }_0^1(\frac{x}{2}-\frac{\frac{x}{2}}{1+2x})dx\right]=\frac{1}{2}log3-\frac{1}{2}{\int }_0^{1}xdx+\frac{1}{2}{\int }_0^1\frac{x}{1+2x}dx=\frac{1}{2}log3-\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^1+\frac{1}{2}{\int }_0^1\frac{\frac{1}{2}(2x+1-1)}{(2x+1)}dx=\frac{1}{2}log3-\frac{1}{2}[\frac{1}{2}-0]+\frac{1}{4}{\int }_0^{1}dx-\frac{1}{4}{\int }_0^1\frac{1}{1+2x}dx=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}[x{]}_0^1-\frac{1}{8}\left[log|\right(1+2x)|{]}_0^1=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}[log3-log1]=\frac{2}{2}-\frac{1}{8}log3=1-\frac{1}{8}log3$