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

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Answer 1 · 2016-08-20T22:26:42

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

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 $ln(1+x)$ :

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

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