1. ## Integration

Def Int from 4-9 (x^1/2 + x^-1/2)^2 dx

2. Originally Posted by bgonzal8
Def Int from 4-9 (x^1/2 + x^-1/2)^2 dx
first expand the brackets. using, of course, $\displaystyle (a + b)^2 = a^2 + 2ab + b^2$

Then use the power rule for integrals to integrate, then continue with the fundamental theorem of calculus.

do ou think you can do the problem now?

3. I did do that and I still got stuck.
I got ... the def. integral from 4-9 of (2x^1/4 + 2x^-1/4)dx so that equals

(8x^5/4)/5 +3/(8x^3/4) + C. The main reason for my confusion is that the question is in the section of my textbook dedicated to Ln(x) so I thought that it was supposed to be part of the answer.

4. $\displaystyle (x^{\frac{1}{2}}$$\displaystyle +x^{\frac{-1}{2}})^2=(\sqrt{x}-\frac{1}{\sqrt{x}})^2$

You have expanded out the x^1/2 terms incorrectly. You will indeed end up with a log term. Think about what $\displaystyle (\sqrt{x})^2$ is.

5. I see that that would make the radical disappear but when I expand it with foil I don't get that as any of the terms. i tried it again and I'm getting x^1/4 +2(sqrt(x))/(sqrt(x)) +x^1/4. And I dont think that's right.

6. Originally Posted by bgonzal8
I see that that would make the radical disappear but when I expand it with foil I don't get that as any of the terms. i tried it again and I'm getting x^1/4 +2(sqrt(x))/(sqrt(x)) +x^1/4. And I dont think that's right.
$\displaystyle (x^{1/2} + x^{-1/2})^2 = (x^{1/2})^2 + 2x^{1/2}x^{-1/2} + (x^{-1/2})^2 = x + 2 + x^{-1}$

7. Ok, so $\displaystyle (\sqrt{x})^2= x$ and $\displaystyle \sqrt{x}=x^{\frac{1}{2}}$.

Try writing your function out as $\displaystyle (\sqrt{x} +\frac{1}{\sqrt{x}})^2$, rather than$\displaystyle ( x^{\frac{1}{2}}+x^{\frac{-1}{2}})^2$

8. Oh jeeze. Thanks you very much. It's always the algebra that gets me.