Integral of $e^{x^3+x^2-1}(3x^4+2x^3+2x)$

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.

Hint: Try to find a primitive of the form $\exp(x^3+x^2-1)Q(x)$, where $Q(x)$ is a polynomial.

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 $\exp(x^3+x^2-1)x^2+c$