Finding the integral using the integration tables. Is this correct? I am trying to find this integral:
$$\int_0^1 x^4*e^{-x}dx$$
I see that:

So in my example, $u = x$, $n = 4$, $a = -1$
So the integral is = $$\frac{1}{-1} x^4*e^{-x} - \frac{4}{-1} \int x^3e^{-x}du + C$$
Is this so far the right track? I just keep going right and reapply the formula?
How does one derive entry 97 anyway? Can someone point me to the derivation?
 A: Yes, you keep reapplying the formula step by step until you reach the situation which is described in 96. 
Concerning the derivation of 97, just use integration by parts. Remember the following: If $f,g : [a,b] \longrightarrow \mathbb{R}$ are continuously differentiable functions, we have
$$ \int_a^b f' g = fg\big\vert_a^b - \int_a^b fg'.$$
Putting $f'(u)=\mathrm{e}^{au}$ and $g(u)=u^n$ in 97, we have (assuming $a \neq 0$)
$$ \int \underbrace{\mathrm{e}^{au}}_{f'} \underbrace{u^n}_{g} \ \mathrm{d}u = \underbrace{\frac{1}{a} \mathrm{e}^{au}}_{f} \underbrace{u^n}_{g}- \int \underbrace{\frac{1}{a}\mathrm{e}^{au}}_{f} \underbrace{n u^{n-1}}_{g'} \ \mathrm{d}u = \frac{1}{a}u^n \mathrm{e}^{au} - \frac{n}{a} \int u^{n-1}\mathrm{e}^{au} \ \mathrm{d}u.$$
A: $$I_n=\int_0^1x^ne^{-ax}dx$$
use integration by parts:
$u=x^n\to u'=nx^{n-1}$ and $v'=e^{-ax}\to v=-e^{-ax}/a$ which will give:
$$I_n=\left[\frac{-x^ne^{-ax}}{a}\right]_0^1-\int_0^1\frac{-nx^{n-1}e^{-ax}}{a}dx$$
$$I_n=-\frac{e^{-a}}{a}+\frac{n}{a}\int_0^1x^{n-1}e^{-ax}dx$$
$$I_n=-\frac{e^{-a}}{a}+\frac naI_{n-1}$$
