# Thread: A more difficult integral, general power formula

1. ## A more difficult integral, general power formula

Question: Integrate 28-1-16

(int) 0.8((3 + 2 ln(u))^3) du/u

My attempt:
u = (3 + 2 ln u)
du = 2(1/u) = 2/u

My guess at this is to divide the whole integral by 2 now since it has been multiplied by 2 in the derivative.

resulting in this:

0.4[3 + 2 ln u]/4

I used a Ti-89 and got a very weird answer. Any clues as to what went wrong? thanks.

2. ## Re: A more difficult integral, general power formula

Hello, togo!

You are using $u$ for two different quantities.

$\displaystyle\int 0.8(3 + 2\ln u)^3\,\frac{du}{u}$

We have: . $0.8\int(3+2\ln u)^3\,\frac{du}{u}$

Let $v \:=\:3+2\ln u \quad\Rightarrow\quad dv \:=\:2\,\frac{du}{u} \quad\Rightarrow\quad \frac{du}{u} \:=\:\tfrac{1}{2}dv$

Substitute: . $0.8\int v^3\left(\tfrac{1}{2}dv\right) \;=\; 0.4\int v^3\,du \;=\;0.1v^4+C$

Back-substitute: . $0.1(3 + 2\ln x)^4 + C$

3. ## Re: A more difficult integral, general power formula

thats cool thanks