# Integral question

1. Jan 26, 2010

### Joy09

1. The problem statement, all variables and given/known data

Evaluate the integral:
the integral of 9r4^r dr

2. Relevant equations

integral of f(x)g'(x)dx = f(x)g(x) - integral of f'(x)g(x)dx

integral of udv = uv - integral vdu

u = f(x), dv = g'(x) dx

3. The attempt at a solution

I first started by pull the 9 out to the front:
9 integral of r4^r dr
I then set u=r, du=dr, dv=4^r, v=(4^r)/ln(4)
I used the formula: integral of udv = uv - integral vdu
and got: the integral of 9r4^r dr = [(r4^r)/ln4] - the inegral of (4^r)/ln4 dr
I then pulled the 1/ln4 out of the last part and got:
the integral of 9r4^r dr = [(r4^r)/ln4] - 1/ln4[the inegral of 4^r dr]

I was also able to get the anti-derivative of the [the inegral of 4^r dr] as (1/ln(4)) 4^r
I got stuck here, i don't know how to put all of it back together, help please?

2. Jan 27, 2010

### vela

Staff Emeritus
You did all the hard stuff already, though you forgot to multiply the 9 through.

This is what you said you have so far:

$$\int 9r4^rdr=9\left(\frac{r4^r}{\log 4}-\frac{4^r}{(\log 4)^2}\right)$$

Aren't you done except for tacking on +C to the end?

3. Jan 27, 2010

### Joy09

yes, but does it needs to be simplified? how do you do that?
my assignment is online and it keeps on telling me that i'm getting the wrong answer.