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

Explanation

`I=int_{0}^{1} x log (1+2 x) d x`
`
=[log (1+2 x) frac{x^{2}}{2}]_{0}^{1}-int frac{1}{1+2 x} cdot 2 cdot frac{x^{2}}{2} d x`
`
=frac{1}{2}[x^{2} log (1+2 x)]_{0}^{1}-int frac{x^{2}}{1+2 x} d x`
`
=frac{1}{2}[log 3-0]-[int_{0}^{1}(frac{x}{2}-frac{frac{x}{2}}{1+2 x}) d x`
`]
=frac{1}{2} log 3-frac{1}{2} int_{0}^{1} {xdx}+frac{1}{2} int_{0}^{1} frac{{x}}{1+2 {x}} {dx}`
`
=frac{1}{2} log 3-frac{1}{2}[frac{x^{2}}{2}]_{0}^{1}+frac{1}{2} int_{0}^{1} frac{frac{1}{2}(2 x+1-1)}{(2 x+1)} d x`
`
=frac{1}{2} log 3-frac{1}{2}[frac{1}{2}-0]+frac{1}{4} int_{0}^{1} {dx}-frac{1}{4} int_{0}^{1} frac{1}{1+2 {x}} {dx}`
`
.=frac{1}{2} log 3-frac{1}{4}+frac{1}{4}[x] 6-frac{1}{8}[log |(1+2 x)|]} `
`
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