Evaluate the following integrals: _ 0 ^1x (1+2x)dx

Question

Evaluate the following integrals:$\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$

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Answer

Let $\text{I}=\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$Apply integral by part
$\text{I}=\Big[\log(1+2\text{x})\frac{\text{x}^2}{2}\Big]^{1}_0-\int_{0}^\limits{1}\Big(\frac{2}{1+2\text{x}}\Big)\times\frac{\text{x}^2}{2}\text{ dx}$
$=\frac{1}{2}\big(\log3-0\big)-\int_{0}^\limits{1}\frac{\text{x}^2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{4\text{x}^2-1+1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{(2\text{x}+1)(2\text{x}-1)}{1+2\text{x}}\text{ dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}(2\text{x}-1)\text{dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\bigg[\frac{1}{4}\times\frac{(2\text{x}-1)^2}{2\times2}\bigg]^1_0-\bigg[\frac{1}{4}\times\frac{\log(1+2\text{x})}{2}\bigg]^1_0$
$=\frac{1}{2}\log3-\frac{1}{16}(1-1)-\frac{1}{8}\big(\log3-\log1\big)$
$=\frac{1}{2}\log3-0-\frac{1}{8}\log3$ $(\log1=0)$
$=\frac{3}{8}\log3$

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