Question

What is the indefinite integral of `ln(1+x)`?

Answer 1

`(x+1)ln(1+x)-x+C`

Explanation:

We have:

`I=intln(1+x)dx`

We will use integration by parts, which takes the form:

`intudv=uv-intvdu`

So, for `intln(1+x)dx`, let:

`{(u=ln(1+x)" "=>" "du=1/(1+x)dx),(dv=dx" "=>" "v=x):}`

Fitting this into the integration by parts formula:

`I=xln(1+x)-intx/(1+x)dx`

In integrating the second bit, you could long divide, but this is simpler:

`I=xln(1+x)-int(1+x-1)/(1+x)dx`

`I=xln(1+x)-int((1+x)/(1+x)-1/(1+x))dx`

`I=xln(1+x)-int(1-1/(1+x))dx`

`I=xln(1+x)-intdx+int1/(1+x)dx`

Both of these are fairly simple integrals:

`I=xln(1+x)-x+ln(1+x)+C`

Factoring `ln(1+x)`:

`I=(x+1)ln(1+x)-x+C`