Question

How do you integrate
`\int\limits ln({x+1}) \, dx` using integration by parts?

Thank you!

Answer 1

`\int \ln (x+1) \, dx=x\cdot \ln (x+1)-x+\ln (x+1)+C`

Explanation:

Use integration by parts formula:

`\int u \, dv=uv-\int v \, du`

For the integral

`\int \ln(x+1) \, dx,`

`u=\ln (x+1)\\ \\dv=dx,`

then

`du=d(\ln (x+1))=(1)/(x+1)\ dx\\ \\v=x`

and

`\int \ln(x+1) \, dx\\ \\=x\cdot \ln (x+1)-\int x\cdot (1)/(x+1) \, dx\\ \\= x\cdot \ln (x+1)-\int (x+1-1)/(x+1) \, dx\\ \\=x\cdot \ln (x+1)-\int \left(1-(1)/(x+1)\right) \, dx\\ \\=x\cdot \ln (x+1)-\int \, dx +\int (1)/(x+1) \, dx\\ \\=x\cdot \ln (x+1)-x+\ln (x+1)+C`