Integration by parts $3n$ times.

Question

I found the equation below in Repeated integration by parts of a definite integral

\begin{align} \int_a^b f^{(n)}(x)g(x) dx = (-1)^{(n)}\int_a^b f(x)g^{(n)}(x) dx \end{align}

which is true if $\left.f^{(k)}(x)g^{(n-k)}(x)\right|_a^b=0 \: \forall k \in [0:n-1]$

So I want to test on the integral below: \begin{align} \int_0^1 \frac{d^{3n}}{dx^{3n}} (x^3-x^4)^n e^x dx \end{align}

here $f(x) = (x^3-x^4)^n$ and $g(x)=e^x$.

Is the following correct?

\begin{align} &\int_a^b f^{(3n)}(x)g(x) dx = (-1)^{(3n)}\int_a^b f(x)g^{(3n)}(x) dx\\ &\int_0^1 \frac{d^{3n}}{dx^{3n}} (x^3-x^4)^n e^x dx = (-1)^{(3n)} \int_0^1 (x^3-x^4)^n e^x dx \end{align}

if not, how to proceed?

Answer 1

The formula is applicable because when you differentiate $f$ $k$ times with $k <n$ there will be at least one factor of $x^{3}-x^{4}$ left in each term and $x^{3}-x^{4}=0$ for $x=0$ and $x=1$.