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

Answer 1

Luke Alvarez

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

We have:

#I=intln(1+x)dx#

We'll employ integration by components, which looks like this:

#intudv=uv-intvdu#

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

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

Including this in the formula for integration by parts:

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

You could use long division to integrate the second bit, but this is easier:

#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#

These two integrals are quite basic:

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

Factoring #ln(1+x)#:

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

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Answer 2

Ezekiel Alexander

The indefinite integral of ln(1+x) is ∫ln(1+x) dx = (x + 1) ln(1 + x) - x + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.