Domain of the Function Defined in Terms of a Series 
The domain of the function $f(x) = \int (x+2x^2 + 3x^3 + ...)dx$ is...

I just want to find out whether the method I used to arrive at the solution is valid. In finding the domain, I first noted that $x+2x^2 + 3x^3 + ... = \sum_{n=1}^{\infty} nx^n$, which is just the derivative of the geometric series, which has the domain (interval of convergence) of $(-1,1)$, and diverges outside the interval. Hence, the domain of $f$ is $(-1,1)$. 
This is the correct answer. However, the solution in the back of the books seems unnecessarily complex. First, the author computes the integral, then performs the ratio test on the terms in the series, which gives him an interval of convergence of $(-1,1)$, and then proceeds to check whether $f$ converges at the endpoints. 
Keep in mind this is for the GRE, so I am looking for the most efficient method while retaining some rigor. 
 A: First of all, when you said that

I first noted that $x+2x^2+3x^3+\ldots=\sum_{n=1}^{\infty}nx^n$, which is just the derivative of the geometric series,

you effectively integrated the given series. You were able to do it quickly by observation, but you integrated nevertheless. So your first step is the same as in the book's solution!
Next, you're right that for the geometric series we already know that its interval of convergence is $(-1,1)$. Making this observation is indeed more efficient than applying the Ratio Test — which is perfectly fine as well, but certainly is more time consuming than simply alluding to a known fact.
But then you make a mistake when you claimed that

Hence, the domain of $f$ is $(−1,1)$.

Termwise integration or differentiation of power series preserves their open intervals of convergence, but the behavior at the endpoints may change. So knowing that the original series (the geometric series in this example) converges on $(-1,1)$, you can NOT immediately deduce that its termwise integral series converges on $(-1,1)$ as well. Its interval of convergence is still from $-1$ to $1$, but we don't know yet whether the endpoints are included or not. That's why the endpoints have to be examined now. It's only a lucky coincidence that in this example the interval is again $(-1,1)$.
