Integral of $e^{x^3+x^2-1}(3x^4+2x^3+2x)$ I understand that this looks exactly like a "Do my Homework" kinda question but trust me, I've spent hours(I won't go into detail as it's off topic).
Note: I'm a High School student, our teacher gave this question as a challenge.
I'm struggling with

$$\int e^{x^3+x^2-1}(3x^4+2x^3+2x)\ dx$$ 

My Progress: I tried finding integral by parts, and found that$$\int e^{x^3+x^2-1}\ dx$$
was the only trouble maker(for now).
So, I tried finding it's integral then gave up and tried using a calculator to see what I missed.
The website said: That it's antiderivative is not elementary. I didn't even know what that means.
New Approach: Now, I went on to trying to plot the graph of $$ e^{x^3+x^2-1}$$  to see if I could related it to $$\int_{-a}^x e^{x^3+x^2-1}\ dx$$ and say whether $$\int e^{x^3+x^2-1}\ dx$$ exists or not.
 I was hoping for some discontinuity in the graph of the definite integral but I didn't seem to find any.
Note: I drew the graph of the definite integral through observation and intuition, I don't think there was any other method.
So is there anyway of helping me?

You can go ahead with integration by parts but taking some time for careful inspection can lead you to a quick answer.

$$\int e^{\color{blue}{x^3+x^2-1}}(3x^4+2x^3+2x)\ dx$$ 

Notice that $\color{blue}{\left( x^3+x^2-1 \right)}' = \color{red}{3x^2+2x}$ and that $3x^4+2x^3 = x^2 \color{red}{\left( 3x^2+2x \right)}$, so you have:
$$\begin{align}
\left(3x^4+2x^3+2x\right)e^{x^3+x^2-1} &
 = x^2\color{red}{\left( 3x^2+2x \right)}e^{\color{blue}{x^3+x^2-1}}+\color{green}{2x}e^{x^3+x^2-1} \\[6pt]
& = x^2\color{blue}{\left( x^3+x^2-1 \right)}'e^{\color{blue}{x^3+x^2-1}}+\color{green}{\left( x^2 \right)'}e^{x^3+x^2-1} \\[6pt]
& = x^2\color{blue}{\left(e^{\color{blue}{x^3+x^2-1}} \right)}'+\color{green}{\left( x^2 \right)'}e^{x^3+x^2-1} \\[6pt]
& = \left( x^2e^{x^3+x^2-1} \right)'
\end{align}$$
where you recognize the product rule for derivatives in the last step.
