Let $I = \int_0^1 x \log (1 + 2x)dx$


$ = \left[ {\log (1 + 2x)\frac{{{x^2}}}{2}} \right]_0^1 - \int {\frac{1}{{1 + 2x}}} \cdot 2 \cdot \frac{{{x^2}}}{2}dx$

$ = \frac{1}{2}\left[ {{x^2}\log (1 + 2x)} \right]_0^1 - \int {\frac{{{x^2}}}{{1 + 2x}}} dx$

$ = \frac{1}{2}[\log 3 - 0] - \left[ {\int_0^1 {\left( {\frac{x}{2} - \frac{{\frac{x}{2}}}{{1 + 2x}}} \right)} dx} \right]$

$ = \frac{1}{2}\log 3 - \frac{1}{2}\int_0^1 x dx + \frac{1}{2}\int_0^1 {\frac{x}{{1 + 2x}}} dx$

$ = \frac{1}{2}[\log 3 - 0] - \left[ {\int_0^1 {\left( {\frac{x}{2} - \frac{{\frac{x}{2}}}{{1 + 2x}}} \right)} dx} \right]$

$ = \frac{1}{2}\log 3 - \frac{1}{2}\int_0^1 x dx + \frac{1}{2}\int_0^1 {\frac{x}{{1 + 2x}}} dx$

$ = \frac{1}{2}\log 3 - \frac{1}{2}\left[ {\frac{{{x^2}}}{2}} \right]_0^1 + \frac{1}{2}\int_0^1 {\frac{{\frac{1}{2}(2x + 1 - 1)}}{{(2x + 1)}}} dx$

$ = \frac{1}{2}\log 3 - \frac{1}{2}\left[ {\frac{1}{2} - 0} \right] + \frac{1}{4}\int_0^1 d x - \frac{1}{4}\int_0^1 {\frac{1}{{1 + 2x}}} dx$

$ = \frac{1}{2}\log 3 - \frac{1}{4} + \frac{1}{4}[x]_0^1 - \frac{1}{8}[\log |(1 + 2x)|]_0^1$
$ = \frac{1}{2}\log 3 - \frac{1}{4} + \frac{1}{4} - \frac{1}{8}[\log 3 - \log 1]$

$ = \frac{1}{2}\log 3 - \frac{1}{8}\log 3$

$ = \frac{3}{8}\log 3$

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