Using u-subsitution to integrate $ \int_{0}^{1} e^{3 x^{3}-2 x}\left(9 x^{2}-2\right) d x $ Question: Use the u-subsitution to integrate $ \int_{0}^{1} e^{3 x^{3}-2 x}\left(9 x^{2}-2\right) d x $

So far my working is as follows
\begin{equation}
u=3 x^{3}-2 x \Longrightarrow \frac{d u}{d x}=9 x^{2}-2
\end{equation}
Step two:
\begin{equation}
\frac{d x}{d u}=\frac{1}{9 x^{2}-2}
\end{equation}
Substituting u in
\begin{equation}
\int_{0}^{1} e^{u}\left(9 x^{2}-2\right) d x
\end{equation}
For definite integration using substitution what changes occur as compared to indefinite u substitution?
EDIT:
After looking at this question I believe I am close to the answer
\begin{equation}
\int_{0}^{1} e^{4}\left(9 x^{2}-2\right) \frac{1}{9 x^{2}-2} d x
\end{equation}
Canceling them out we get;
\begin{equation}
\int_{0}^{1} e^{u} d u
\end{equation}
when x=1, our new limit is 7 when x=0 our new limit is -2
 A: \begin{equation}
\frac{d x}{d u}=\frac{1}{9 x^{2}-2},
\end{equation}
We are allowed to multiply both sides by $du$ to get:
\begin{equation}
d x=\frac{1}{9 x^{2}-2} du.
\end{equation}
Substitute this into \begin{equation}
\int_{x=0}^{x=1} e^{u}\left(9 x^{2}-2\right) d x.
\end{equation}
But then you must get the limits $x=1$ and $x=0$ also in terms of $u.$
So you have to replace $x=1$ with $u=?$ [What is $u$ when $x=1$?]
And similar for $x=0.$
Only once everything is in terms of $u$ are you allowed to evaluate the integral.
A: Your expression after u-substitution is just $$\int_0^1 e^u du=e^{3x^2-2x}\Bigg |^1_0$$ You can do the rest.
A: Upon inspection you can write
$$
\int_0^1e^{3x^3-2x}(9x^2-2)\,\mathrm dx=\int_0^1e^{3x^3-2x}\,\mathrm d(3x^3-2x).
$$
Since we are now integrating with respect to the quantity in the exponent, the result follows immediately
$$
\int_0^1e^{3x^3-2x}\,\mathrm d(3x^3-2x)=e^{3x^3-2x}|_{x=0}^1=e-1.
$$
Admittedly, this is just $u$-substitution in disguise.
A: So from my workings I computed
\begin{equation}
\left(e^{u}\right)_{0}^{1}=e^{1}-e^{0}=e-1
\end{equation}
This is from
\begin{equation}
\int_{0}^{1} e^{u} d u
\end{equation}
Therefore giving our final answer
\begin{equation}
\left(e^{u}\right)_{0}^{1}=e^{1}-e^{0}=e-1
\end{equation}
