Finding the interval of convergence from the integral of a power series 
Let
$$\sum_{n=0}^∞ (-1)^n \bigg(\frac{x}{4}\bigg)^n$$
Find the series and interval of convergence for the integral from $0$ to $x$ of $f(t)dt.$

I'm a bit confused. Do I find the sum of the series, which I found to be
$$-\frac{x}{(4+x)}$$
and find the integral of that from $0$ to $x$? I was left with $4\ln (x+4)-x-4\ln(4)$. Or do I take the integral of the original function immediately, replacing the $x$ with $t$ leaving me with
$$\frac{(-1)^n \cdot x^{(n+1)}}{ (n+1)\cdot 4^n}$$
using this to find the interval of convergence?
Edit: got it! thank you so much everyone!
 A: Recall geometric series $$ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x},\quad x\in(-1,1).$$
Your function $f(x)$ can be written as geometric series
$$ f(x) = \sum_{n=0}^{\infty} (-1)^n(\frac{x}{4})^n = \sum_{n=0}^{\infty} (-\frac{x}{4})^n = \frac{1}{1-(-x/4)} = \frac{4}{x+4}, x\in(-4,4).$$
So your summation is not correct.
To represent the integral of $f(t)$ as power series, we can simply replace it by using power series. (no need to find summation of $f(x)$) Then we switch the order of integration and summation (we must be careful when we change the order but we can do it for your questions), then it becomes to
$$\int_{0}^{x}f(t)dt = \int_{0}^{x}\sum_{n=0}^{\infty} (-\frac{t}{4})^n dt = \sum_{n=0}^{\infty}\int_{0}^{x} (-\frac{t}{4})^n dt = \sum_{n=0}^{\infty}(-1)^n\frac{1}{n+1}\frac{x^{n+1}}{4^n}. $$
Integration doesn't change the radius of convergence which is $R=4$, but you should check endpoints $x=-4$ and $x=4$ to get interval of convergence. It's possible that the power series is convergent at both endpoints, only one endpoints or is divergent at both endpoints.
A: hint
If $ |a|<1$, make the difference between $$a+a^2+a^3+...= a\frac{1}{1-a}$$
and
$$1+a+a^2+a^3+...= \frac{1}{1-a}$$
In your question, If
$$|\frac x4|<1 \text{ or } |x|<4$$
$$f(x)=\sum_{k=0}^{+\infty}(-\frac x4)^k=\frac{1}{1+\frac x4}$$
$$\int_0^xf(t)dt=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)4^k}x^{k+1}$$
