Question

What is the integral from 0 to infinity of (x^n)(e^(-4x)) dx

Answer 1

Ignore the limits of integration for a sec, you can deal with those at the end.\[\huge \int\limits (x^n) e^{-4x}dx\]Applying `Integration by Parts`:
\[\large \;\;\;u=x^n \qquad \qquad \qquad \quad \;\;dv=e^{-4x}dx\]\[\large du=nx^{n-1}\qquad \qquad \qquad v=-\frac{1}{4}e^{-4x}\]

Plugging in the pieces gives us,\[\large -\frac{1}{4}(x^n)e^{-4x}+\frac{n}{4}\int\limits (x^{n-1})e^{-4x}dx\]
We'll have to apply integration by parts again. Our \(\large u\) this time will instead by \(\large x^{n-1}\). Which will give us,\[\large -\frac{1}{4}(x^n)e^{-4x}-\frac{n}{4^2}(x^{n-1})e^{-4x}+\color{orangered}{\frac{n(n-1)}{4^2}}\int\limits (x^{n-2})e^{-4x}dx\]

Answer 2

Do you see a pattern forming yet? :)
If you evaluate the different parts at your upper and lower limits, you should discover something interesting.

As you apply Integration by Parts over and over, you'll notice the fraction on the integral part keeps growing in complexity.
That's the part to pay attention to!