# How do you find the Taylor series of an indefinite integral?

I am given the following problem :-

I know that I have to first find the Taylor series of the polynomial and then integrate each term, however I am having trouble finding the Taylor series before integrating because the derivative of $$\frac{(e^t-e)}{(t-1)}\to \frac{e^t(t-1)-1(e^t-e)}{(t-1)^2}$$ at $$t=1$$ does not exist $$\frac{0}{0}$$. How do I go about finding the Taylor series of $$\frac{(e^t-e)}{(t-1)}?$$

Thanks for any and all help!

• The integrand is only undefined at $x=0$, in a removable way, so its antiderivative is defined everywhere and turns out to be differentiable.
– user65203
Commented Aug 5, 2020 at 14:57

Set $$t=1+u\enspace(u\to 0)\,$$ first. The integrand becomes $$\frac{\mathrm e^t-\mathrm e}{t-1}=\mathrm e\,\frac{\mathrm e^u-1}u.$$ Can you take it from here?
• Why do you want to take the derivative? The numerator has a well-known expansion, and everything simplifies. If you compute the derivative and consider the expansion of the numerator , the terms with order less than $2$ disappear, so it's not indeterminate. Commented Aug 5, 2020 at 15:20