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?
 A: Hint: Try to find a primitive of the form $\exp(x^3+x^2-1)Q(x)$, where $Q(x)$ is a polynomial.
A: Assume the solution to be of the form $\exp(x^3+x^2-1)Q(x)$ and differentiate this expression to get $\exp(x^3+x^2-1)\left[(3x^2+2x)Q(x)+Q'(x)\right]$. Compare this with $\exp(x^3+x^2-1)\left[3x^4+2x^3+2x\right]$ to obtain the ODE:
$$(3x^2+2x)Q(x)+Q'(x)=3x^4+2x^3+2x.$$
The particular solution is $x^2$, you could also solve this ODE with separation of variables and variation of parameters to get $$Q(x)=x^2+c_1\exp(-x^3-x^2).$$ Setting $c_1=0$ will result in the particular solution $x^2$.
So, the integral is given by $\exp(x^3+x^2-1)x^2+c$
A: 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.
A: Let $$I = \int e^{x^3+x^2-1}\bigg(3x^4+2x^3+2x\bigg)dx = \int e^{x^3+x^2-1}\cdot x^2\bigg(3x^2+2x+2x^{-1}\bigg)dx$$
So $$I = \int e^{x^3+x^2+2\ln x-1}\bigg(3x^2+2x+2x^{-1}\bigg)dx$$
Put $x^3+x^2+2\ln x-1 = t\;,$ Then $(3x^2+2x+2x^{-1})dx = dt$
So $$I = \int e^t dt = e^t+\mathcal{C} = e^{x^3+x^2+2\ln x-1}+\mathcal{C} = x^2\cdot e^{x^3+x^2-1}+\mathcal{C}$$
A: Well, we would like to have in our integral something like the derivative of $x^3-x^2-1$, which is $3x^2+2x$. So, let us just divide $3x^4+2x^3+2x$ by $3x^2+2x$. We will get:
$$3x^4+2x^3+2x=x^2(3x^2+2x)+2x$$
So, let's plug it in our integral:
$$\begin{align*}\int e^{x^3+x^2-1}(3x^4+2x^3+2x)dx=&\int e^{x^3+x^2-1}\left(x^2(3x^2+2x)+2x\right)dx=\\
=&\int x^2e^{x^3+x^2-1}(3x^2+2x)+2xe^{x^3+x^2-1}dx=\\
=&\int x^2\left(e^{x^3+x^2-1}\right)'+(x^2)'e^{x^3+x^2-1}dx=\\
=&\int \left(x^2e^{x^3+x^2-1}\right)'dx=\\
=&x^2e^{x^3+x^2-1}+c,\ c\in\mathbb{R}
\end{align*}$$
Hope this helped! :)
