Let I= _ 0 ^ 1 x (1+2x)dx = [ (1+2x) x^2 2 ]_ 0 ^ 1 - 1 1+2x 2 x^2 2 dx =1 2 [x^2 (1+2x) ]_ 0 ^ 1 - x^2 1+2x dx =1 2 [ 3-0]- [ _ 0 ^ 1 ( x 2 - x 2 1+2x )dx ] =1 2 3-1 2 ^ 1 _ 0 x dx+1 2 ^ 1 _ 0 x 1+2x dx =1 2 3-1 2 [ x^2 2 ]^ 1 _ 0 +1 2 ^ 1 _ 0 1 2 (2x+1-1) (2x+1) dx =1 2 3-1 2 [1 2 -0 ]+1 4 ^1_0dx-1 4 ^1_01 1+2x dx =1 2 3-1 4 +1 4 [x ]^ 1 _ 0 -1 8 [ |(1+2x)| ]^ 1 _ 0 =1 2 3-1 4 +1 4 -1 8 [ 3- 1 ] =1 2 3-1 8 3 =3 8 3

_ 0 ^ 1 x (1+2x)dx

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$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$

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Answer

Let $\text{I}=\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
$=\Bigg[\log(1+2\text{x})\frac{\text{x}^2}{2}\Bigg]_{0}^{1}-\int\frac{1}{1+2\text{x}}\cdot2\cdot\frac{\text{x}^2}{2}\text{dx}$
$=\frac{1}{2}\big[\text{x}^2\log(1+2\text{x})\big]_{0}^{1}-\int\frac{\text{x}^2}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}[\log3-0]-\Bigg[\int\limits_{0}^{1}\bigg(\frac{\text{x}}{2}-\frac{\frac{\text{x}}{2}}{1+2\text{x}}\bigg)\text{dx}\Bigg]$
$=\frac{1}{2}\log3-\frac{1}{2}\int\limits^{1}_{0}\text{x dx}+\frac{1}{2}\int\limits^{1}_{0}\frac{\text{x}}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{2}\bigg[\frac{\text{x}^2}{2}\bigg]^{1}_{0}+\frac{1}{2}\int\limits^{1}_{0}\frac{\frac{1}{2}(2\text{x}+1-1)}{(2\text{x}+1)}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{2}\Big[\frac{1}{2}-0\Big]+\frac{1}{4}\int^1_0\text{dx}-\frac{1}{4}\int^1_0\frac{1}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{4}+\frac{1}{4}\big[\text{x}\big]^{1}_{0}-\frac{1}{8}\big[\log|(1+2\text{x})|\big]^{1}_{0}$
$=\frac{1}{2}\log3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\big[\log3-\log1\big]$
$=\frac{1}{2}\log3-\frac{1}{8}\log3$
$=\frac{3}{8}\log3$

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