Answer:
[tex]\int \ln (x+1) \, dx=x\cdot \ln (x+1)-x+\ln (x+1)+C[/tex]
Step-by-step explanation:
Use integration by parts formula:
[tex]\int u \, dv=uv-\int v \, du[/tex]
For the integral
[tex]\int \ln(x+1) \, dx,[/tex]
[tex]u=\ln (x+1)\\ \\dv=dx,[/tex]
then
[tex]du=d(\ln (x+1))=\dfrac{1}{x+1}\ dx\\ \\v=x[/tex]
and
[tex]\int \ln(x+1) \, dx\\ \\=x\cdot \ln (x+1)-\int x\cdot \dfrac{1}{x+1} \, dx\\ \\= x\cdot \ln (x+1)-\int \dfrac{x+1-1}{x+1} \, dx\\ \\=x\cdot \ln (x+1)-\int \left(1-\dfrac{1}{x+1}\right) \, dx\\ \\=x\cdot \ln (x+1)-\int \, dx +\int \dfrac{1}{x+1} \, dx\\ \\=x\cdot \ln (x+1)-x+\ln (x+1)+C[/tex]
