What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor series form.
I mean:
$$\int _0 \frac{d}{dx}\sum_{n=1}^\infty (n^2+n^3)x^{n-1}dx=\int_0 \sum_{n=1}^\infty \frac{n^2+n^3}{n-1}x^{n-2}dx$$
But it doesn't seem to work in this exercise.

Another way to solve it.... 

First: 
Note that 
$$\frac{d^2}{dx^2}\Bigl(nx^{n+1}\Bigr)=n^2(n+1)x^{n-1}=(n^2+n^3)x^{n-1}\qquad(1)$$
we will use this later.

Second: 
We know that
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\qquad|x|<1$$
Taking the derivative respect to $x$, we have:
$$\begin{array}{rcl}
\frac{d}{dx}\Bigg(\sum_{n=0}^{\infty}x^n\Bigg)&=&\frac{d}{dx}\Bigg(\frac{1}{1-x}\Bigg)\\
\sum_{n=0}^{\infty}\frac{d}{dx}x^n&=&\frac{1}{(1-x)^2}\\
\sum_{n=1}^{\infty}nx^{n-1}&=&\frac{1}{(1-x)^2}\\
\sum_{n=0}^{\infty}nx^{n-1}&=&\frac{1}{(1-x)^2}\qquad(2)
\end{array}$$
where in the LHS of $(2)$ I summed a $0$, that is equivalent to the term for $n=0$.

Third:
Take $(2)$ and multiply by $x^2$:
$$\sum_{n=0}^{\infty}nx^{n+1}=\frac{x^2}{(1-x)^2}\qquad|x|<1\qquad(3)$$
Now we have the general term $nx^{n+1}$ in the sum, so if we derive respect to $x$ two times, we will have our function because of the relation of $(1)$. Then:
$$\begin{array}{rcl}
\frac{d^2}{dx^2}\Bigg(\sum_{n=0}^{\infty}nx^{n+1}\Bigg)&=&\frac{d^2}{dx^2}\Bigg(\frac{x^2}{(1-x)^2}\Bigg)\\
\sum_{n=1}^{\infty}\frac{d^2}{dx^2}\Big(nx^{n+1}\Big)&=&\frac{d}{dx}\Bigg(\frac{-2x}{(1-x)^3}\Bigg)\\
\sum_{n=1}^{\infty}(n^2+n^3)x^{n-1}&=&\frac{2(2x+1)}{(1-x)^4}\\
\end{array}$$
Note we had to drop the constant term that happens at $n=0$ for the 1st derivative of the series (second line).
