Question

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

Answer 1

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

Applying integration by parts, we have

$$I=\log (1+2x){\frac{x^2}{2}}_0^1-{\int }_0^1\left(\frac{2}{1+2x}\right)\times \frac{x^2}{2}dx$$
$$=\frac{1}{2}(\log 3-0)-{\int }_0^1\frac{x^2}{1+2x}dx$$
$$=\frac{1}{2}\log 3-\frac{1}{4}{\int }_0^1\frac{4x^2-1+1}{1+2x}dx$$
$$=\frac{1}{2}\log 3-\frac{1}{4}{\int }_0^1\frac{(2x+1)(2x-1)}{1+2x}dx-\frac{1}{4}{\int }_0^1\frac{1}{1+2x}dx$$
$$=\frac{1}{2}\log 3-\frac{1}{4}{\int }_0^1(2x-1)dx-\frac{1}{4}{\int }_0^1\frac{1}{1+2x}dx$$

$$={\frac{1}{2}\log 3-\frac{1}{4}\times \frac{{(2x-1)}^2}{2\times 2}|}_0^1-{\frac{1}{4}\times \frac{\log (1+2x)}{2}|}_0^1$$
$$=\frac{1}{2}\log 3-\frac{1}{16}(1-1)-\frac{1}{8}(\log 3-\log 1)$$
$$=\frac{1}{2}\log 3-0-\frac{1}{8}\log 3....................(\log 1=0)$$
$$=\frac{3}{8}\log 3$$