\(\int \log (1+x)^{1+x} \mathrm{~d} x=\)

\(\begin{aligned} & \text {Let } \mathrm{I}=\int \log (1+x)^{1+x} \mathrm{~d} x \\ & \mathrm{I}=\int(1+x) \cdot \log (1+x) \mathrm{d} x \\ & \text {Put } 1+x=\mathrm{t} \\ \therefore \quad & \mathrm{d} x=\mathrm{dt} \\ \therefore \quad & \mathrm{I}=\int \mathrm{t} \cdot \log \mathrm{t} \cdot \mathrm{dt}\end{aligned}\)
\(\begin{aligned} & =\log \mathrm{t} \int \mathrm{t} \mathrm{~dt}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \log \mathrm{t} \cdot \int \mathrm{tdt}\right) \mathrm{dt} \\ & =\log \mathrm{t} \cdot \frac{\mathrm{t}^2}{2}-\int\left(\frac{1}{\mathrm{t}} \times \frac{\mathrm{t}^2}{2}\right) \mathrm{dt} \\ & =\log \mathrm{t} \cdot \frac{\mathrm{t}^2}{2}-\frac{1}{2} \int \mathrm{t} d \mathrm{dt} \\ & =\log \mathrm{t} \frac{\mathrm{t}^2}{2}-\frac{\mathrm{t}^2}{4}+\mathrm{c} \\ & =\mathrm{t}^2 \frac{\log \mathrm{t}}{2}-\frac{\mathrm{t}^2}{4}+\mathrm{c} \\ & =\frac{\mathrm{t}^2}{2}\left[\log \mathrm{t}-\frac{1}{2}\right]+\mathrm{c} \\ \mathrm{I} & =\frac{(1+x)^2}{2}\left[\log (1+x)-\frac{1}{2}\right]+\mathrm{c}\end{aligned}\)