# Thread: Indefinite Integral as Power Series

1. ## Indefinite Integral as Power Series

I have no idea how to do this

I have to write this integral:

2. Originally Posted by phack
I have no idea how to do this

I have to write this integral:

x Log[1+n^2]/n^2 + c

unless you really want integral log(1+x^2) / x^2 dx

Then we can write:

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

for |x|<1.

Then:

integral log(1+x^2) / x^2 dx = sum_{n=1 to infnty} integral (-1)^{n+1}x^{2n-1}/n dx

.................= sum_{n=1 to infnty} (-1)^{n+1}x^{2n}/(n (2n-1)

RonL

3. I'm not sure I'm following your steps. Why did you place an x infront in your first step?

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

4. Originally Posted by phack
I'm not sure I'm following your steps. Why did you place an x infront in your first step?

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

Because,

ln(1+x)= x - x^2/2+x^3/3 -... for -1<x<=1

But you have,

ln(1+x^2)

Thus, instead of x you have x^2.

5. Originally Posted by phack
I'm not sure I'm following your steps. Why did you place an x infront in your first step?
Because in your question you have an integral over x, but no x in the
integrand. Therefore the integral is x times whatever you had under
the integral sign (as this was independent of x). Therefore the first step
the question I assumed that you realy wanted to ask.

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...