# evaluate the indefinite integral

• Oct 21st 2010, 08:57 PM
Taurus3
evaluate the indefinite integral
Evaluate the indefinite integral of x(2x+5)^8dx

Alright, so far I've got:

u=2x+5, du=2dx, (1/2)du=dx, x = (u-5)/2
integral of (u-5)/2*u^8*1/2du
1/2integral of (u-5)/2*u^8du
1/2 ((2x+5)-5)/2*((2x+5)^9)/9 + C

But this wasn't the answer...8( Am I doing something wrong?
• Oct 21st 2010, 09:16 PM
Prove It
You would need to use integration by parts...

Let $\displaystyle u = x$ and $\displaystyle dv = (2x + 5)^8$.

Then $\displaystyle du = 1$ and $\displaystyle v = \frac{1}{18}(2x + 5)^9$.

So $\displaystyle \int{x(2x+5)^8\,dx} = \frac{1}{18}x(2x + 5)^9 - \int{\frac{1}{18}(2x + 5)^9\,dx}$

$\displaystyle = \frac{1}{18}x(2x + 5)^9 - \frac{1}{36}\int{2(2x + 5)^9\,dx}$

$\displaystyle = \frac{1}{18}x(2x + 5)^9 - \frac{1}{360}(2x + 5)^{10} + C$.
• Oct 21st 2010, 09:29 PM
tonio
Quote:

Originally Posted by Taurus3
Evaluate the indefinite integral of x(2x+5)^8dx

Alright, so far I've got:

u=2x+5, du=2dx, (1/2)du=dx, x = (u-5)/2
integral of (u-5)/2*u^8*1/2du
1/2integral of (u-5)/2*u^8du
1/2 ((2x+5)-5)/2*((2x+5)^9)/9 + C

But this wasn't the answer...8( Am I doing something wrong?

$\frac{1}{4}\int(u-5)u^8du=\frac{1}{4}\int(u^9-5u^8)du=\frac{1}{4}\left(\frac{u^{10}}{10}-\frac{5u^9}{9}\right)+C\,,\,\,C=$ constant
• Oct 21st 2010, 09:35 PM
Taurus3
• Oct 21st 2010, 09:37 PM
Prove It
Which is what you get after you back substitute $u = 2x + 5$.

My answer is also correct, just written in a different form.
• Oct 21st 2010, 09:39 PM
Educated
Quote:

Originally Posted by Taurus3
I'm assuming it is: $\frac{1}{40}(2x+5)^{10} - \frac{5}{36}(2x+5)^9+C$
$\frac{1}{4}\left(\frac{u^{10}}{10}-\frac{5u^9}{9}\right)+C$