What is the antiderivative of (ln(x+1)x2)?

1 Answer
Lovecraft
Jan 9, 2016

I=−ln(x+1)x+ln|x|−ln|x+1|+c

Explanation:

I=∫ln(x+1)x2dx

Let's say u=ln(x+1) so du=1x+1, and dv=1x2 so v=−1x

I=−ln(x+1)x+∫dxx(x+1)

The latter integral can only be solved with parcial fractions, so we assume there is a sum of fractions

ax+bx+1=1x(x+1)

Where a and b are constants, that is also to say

a(x+1)+bxx(x+1)=1x(x+1)

Note that it means, that, for any value of x

a(x+1)+bx=1

So if we assume x=0,

a(0+1)+b⋅0=1
a=1

And if we assume x=−1

a(−1+1)−b=1
−b=1
b=−1

So, back to the first integral we have

I=−ln(x+1)x+∫(1x−1x+1)dx
I=−ln(x+1)x+∫dxx−∫dxx+1

From there, it's an easy integral

I=−ln(x+1)x+ln|x|−ln|x+1|+c

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