Question

$\underset{0}{\overset{1}{\int }}\log \left(1+x\right)dx$

Answer 1

$$\begin{gathered} {\int }_0^1\log \left(1+x\right)dx \\ ={\int }_0^1\log \left(1+x\right)\times 1dx \\ ={\left[\log \left(1+x\right)x\right]}_0^1-{\int }_0^1\frac{x}{1+x}dx \\ ={\left[\log \left(1+x\right)x\right]}_0^1-{\int }_0^1\left(1-\frac{1}{1+x}\right)dx \\ ={\left[x\log \left(1+x\right)\right]}_0^1-{\left[x-\log \left(1+x\right)\right]}_0^1 \\ =\log 2-1+\log 2 \\ =2\log 2-1 \\ =\log 4-\log e \\ =\log \frac{4}{e} \end{gathered}$$