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

1 Answer
Aug 20, 2016

Answer:

#(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#