More reduction formulae troubles

[NewtonianAlch]

Homework Statement

$\int$x$^{n}$2$^{x}$dx = $\frac{2}{ln 2}$ - $\frac{n}{ln 2}$ $\int$x$^{n-1}$2$^{x}$dx

For n is greater or equal to 1, find $\int$x$^{3}$2$^{x}$dx

This is a definite integral from 0 to 1

The Attempt at a Solution

My first question is, after reducing this once, you are left with x$^{2}$, and attempting to reduce that again means the constant outside the integral will effectively become 0, therefore the constant term outside the integral will be 0 no matter how many more times you reduce it, and after you do the integral, 0 times the integrated result will be zero. What am I not seeing properly here?

Ignoring that and continuing to do the reduction, I get $\frac{1}{ln 2}$*$\frac{1}{ln 2}$ which doesn't seem right to ignore it anyways.

That is to say, when it becomes x$^{1}$, and you apply the formula again, you'll get $\frac{1}{ln 2}$ outside the integral, then you're left with x$^{0}$ which is 1, and effectively you just integrate 2$^{x}$ which becomes $\frac{1}{ln 2}$2$^{x}$, substituting 0 and 1, and doing the maths, you're left with $\frac{1}{ln 2}$*$\frac{1}{ln 2}$ as the answer.

 

[Screwdriver]

I'm pretty sure that the formula is actually this (by using parts):

$$\int x^{n}2^{x}dx=\frac{2^{x} x^{n}}{ln(2)}-\frac{n}{ln(2)}\int x^{n-1}2^{x}dx$$

Applying this once:

$$\int x^{3}2^{x}dx=\frac{2^{x} x^{3}}{ln(2)}-\frac{3}{ln(2)}\int x^{3-1}2^{x}dx$$

http://www.wolframalpha.com/input/?i=integrate+x^%283%29+2^x+dx+%3D+%28%282^x+x^3%29%2Fln%282%29%29+-+%283+%2F+ln%282%29%29+integrate+x^%283+-+1%29+2^x+dx