# Help with integration

• Apr 9th 2008, 08:24 AM
nisse pisse
Help with integration
I supposed to integrate (x^2 + x)^(1/2).
I do it this way

int (x^2 + x)^(1/2) = x(x^2 + x)^(1/2) - 1/2* int (x(2x + 1))/(x^2 + x)^(1/2) = x (x^2 + x)^(1/2) - 1/2((xln(x^2 +x)) - ln(x^2 + x))

IF u now take 2pi ((x = 1) - (x = 0)) u are supposed to get
pi(6(2)^(1/2) - ln(3 + 2(2)^(1/2)) / 4

And i dont get it, so i wonder what im doing wrong.(Headbang)
• Apr 9th 2008, 08:35 AM
ThePerfectHacker
Quote:

Originally Posted by nisse pisse
I supposed to integrate (x^2 + x)^(1/2).

(Say that we are integrating on the positive real line).

$\displaystyle \int (x^2+x)^{1/2}dx = \int \frac{2x(x+1)^{1/2}}{2\sqrt{x}}dx$
Let $\displaystyle t=\sqrt{x}\implies t' = \frac{1}{2\sqrt{x}}$ and so,
$\displaystyle \int 2t^2 \sqrt{t^2+1}dt$

Can you finish this?
• Apr 10th 2008, 02:30 AM
nisse pisse
Now i get 2*(((t^3)/3)*(t^2 + 1)^(1/2)) - (int) (((t^4)/3) * (1/(t^2 + 1)^(1/2)) = 2*(((t^3)/3)*(t^2 + 1)^(1/2)) - ((((t^4)/3) * ln(t + (t^2 + 1)^(1/2))) - (int) (4t/3)*ln(t + (t^2 + 1)^(1/2)))
I dont see where to go from my last step, becuse i dont the int of
ln(t + (t^2 + 1)^(1/2))and there got to be an easier way than finding out.
• Apr 10th 2008, 02:48 AM
angel.white
Quote:

Originally Posted by ThePerfectHacker
(Say that we are integrating on the positive real line).

$\displaystyle \int (x^2+x)^{1/2}dx = \int \frac{2x(x+1)^{1/2}}{2\sqrt{x}}dx$
Let $\displaystyle t=\sqrt{x}\implies t' = \frac{1}{2\sqrt{x}}$ and so,
$\displaystyle \int 2t^2 \sqrt{t^2+1}dt$

Can you finish this?

Wow, I really liked that. That was much more elegant that what I would have done.
• Apr 10th 2008, 12:03 PM
Krizalid
But there's no way to avoid a trig. sub. here. You're gonna have to apply it yes or yes. (Integration by parts might work.)