Taylor Polynomial for Log(1-t) derivation {confused}

I need some help understanding something from my textbook. In the book they are deriving a taylor polynomial approximation for log(1-t), the first thing they do is integrate log(1-t) from 0 to t.

What I don't understand is why are we integrating in the first place?

If I wanted to find the taylor polynomial representation for this function wouldn't I take the derivatives and construct it using the Taylor series formula?

Some help / explanation would be great

• Are you sure they don’t integrate $1/(1-u)$ from 0 to $t$? – TM Gallagher Apr 10 at 17:33
• @TMGallagher positive – Temirzhan Apr 10 at 17:34
• I would have guessed that they observed the following: $$\int_0^t \frac{1}{1-u} \, du = -\ln(1-t)$$ and then use the fact that you know the Taylor series expansion for $1/(1-u)$. – TM Gallagher Apr 10 at 17:37
• @TMGallagher I think I understand what they're doing now.. It's a bit of a weird derivation in my opinion. I also haven't done calculus in a while. Thanks! – Temirzhan Apr 10 at 18:17