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Evaluate : int0^1xlog(1+2x)dx...

Evaluate : `int_0^1xlog(1+2x)dx`

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The correct Answer is:
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Let `I = int_(0)^(1)xlog(1+2x)dx`
`= [log(1+2x)(x^(2))/(2)]_(0)^(1)-int(1)/(1+2x).2.(x^(2))/(2)dx`
` =1/2[x^(2)log(1+2x)]_(0)^(1)-int(x^(2))/(1+2x)dx`
` = 1/2[1log3-0]-[int_(0)^(1)(x/2-(x/2)/(1+2x))dx]`
`= 1/2 log 3 - (1)/(2)int_(0)^(1)xdx+ 1/2 int_(0)^(1)(x)/(1+2x)dx`
`= 1/2log3-1/2[(x^(2))/(2)]_(0)^(1)+1/2int_(0)^(1)(1/2(2x+1-1))/((2x+1))dx`
`= 1/2log3-1/2[1/2-0]+1/4int_(0)^(1)dx - 1/4int_(0)^(1) (1)/(1+2x)dx`
`= 1/2log3-1/4+1/4[x]_(0)^(1)-1/8[log|(1+2x)|]_(0)^(1)`
`= 1/2 log3 - 1/4 + 1/4 - 1/8 [log3-log1]`
`= 1/2 log3 - 1/8 log3`
` = 3/8 log3`
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