# Math experts please come

1. Feb 23, 2007

### rasensuriken

Math experts please come!!

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

Calculus, integration

2. Relevant equations

3. The attempt at a solution

used method:changing variable, by part, trigo etc but no avail. Please help me solve it, thanks.

Last edited: Feb 23, 2007
2. Feb 23, 2007

### gammamcc

SD, Your kidding, right? Didn't your high school math teacher make you do a "time out" whenever misused the square root that way?

Last edited: Feb 23, 2007
3. Feb 23, 2007

### gammamcc

Rewrite the argument x^2-x by "completing the square" . "u" substitution follows.

4. Feb 23, 2007

### ManMonkeyFish

Did you try a U substition with U = X^2 - X and dU = 2x - 1 dx?

5. Feb 23, 2007

### VietDao29

When you meet the problem asking you to integrate something like:
$$\int \frac{dx}{(ax + b) ^ {m} \sqrt{c x ^ 2 + dx + e}}$$, what you should do is to let:
$$ax + b = \frac{1}{t}$$
--------------------
So, here, we go:
$$\int \frac{dx}{(2x + 1) \sqrt{x ^ 2 - x}}$$
For $$\sqrt{x ^ 2 - x}$$ to be valid, x can only be in the interval $$] - \infty ; 0 ] \unity [1 ; + \infty [$$
Let
$$2x + 1 = \frac{1}{t} \quad \mbox{or} \quad x = \frac{1}{2} \left( \frac{1}{t} - 1 \right)$$
$$\Rightarrow 2dx = -\frac{dt}{t ^ 2}$$
The integral becomes:
$$- \int \frac{\frac{dt}{2 t ^ 2}}{\frac{1}{t} \sqrt{ \left[ \frac{1}{2} \left( \frac{1}{t} - 1 \right) \right] ^ 2 - \frac{1}{2} \left( \frac{1}{t} - 1 \right)}}$$

$$= - \int \frac{dt}{2t\sqrt{ \left[ \frac{1}{4} \left( \frac{1}{t ^ 2} - \frac{2}{t} + 1 \right) \right] - \left( \frac{1}{2t} - \frac{1}{2} \right)}}$$

$$= - \int \frac{dt}{2t\sqrt{\frac{1}{4 t ^ 2} - \frac{1}{t} + \frac{3}{4}}}$$

$$= - \int \frac{dt}{t \sqrt{\frac{1}{t ^ 2} - \frac{4}{t} + 3}}$$

$$= \pm \int \frac{dt}{\sqrt{1 - 4t + 3 t ^ 2}}$$
Can you go from here? :)

Last edited: Feb 23, 2007
6. Feb 23, 2007

### gammamcc

Forget what I had written here

Last edited: Feb 23, 2007
7. Feb 23, 2007

### Plastic Photon

Won't work because the integrand has dx/(2x-1) and not (2x-1)dx.

The best approach here is partial fraction decomp.

Let 1=A(2x-1) + B(x^2-x)^1/2

Set x=1 , A=1
Set x=1/2 , B=2i

Then integrate the new functions 1/(x^2-x)^1/2 and 2i/(2x-1)

EDIT: The function 1/(x^2-x)^1/2 can be broken down further into |x^1/2|(x-1)^1/2 or into ((x-x^1/2)(x+x^1/2))^1/2, though the first of these two should look familiar.

Last edited: Feb 23, 2007
8. Feb 23, 2007

### SunGod87

(x^2-x)^(1/2) = x - x^(1/2) is not true. You can check by multiplying the RHS with itself.

9. Feb 23, 2007

### arildno

$$\sqrt{4^{2}-4}=\sqrt{16-4}=\sqrt{12}=2=4-\sqrt{4}$$
Fascinating..

10. Feb 23, 2007

### Schrodinger's Dog

OK my bad. I was thinking of $$\sqrt{x^2} - \sqrt{x}$$

In my defence I had just finished a long and arduous day at work and wasn't really thinking sorry.

I deleted it, still the shame lives on.... I'll go bury myself under a pile of soil and pretend I'm dead

For my next trick I'll prove Fermat's last theorem in a new and exciting way whilst drunk :/

Last edited: Feb 23, 2007
11. Feb 23, 2007

### gammamcc

going with my first reply

Opps, I got fooled. Monkey Man's answer seemed so slick. I now vote for my own approach. It boils down to a constant times integral of
1/(u sqrt(u^2 -a^2))

"complete the square."

I don't trust partial fractions since we don't have a rational function.

12. Feb 23, 2007

### Plastic Photon

Your kidding right? So 1/((2x-1)(x^2-x)^1/2) is not a rational function?

13. Feb 23, 2007

### gammamcc

Nope, It's not. Rational functions are ratios of polynomials.

14. Feb 23, 2007

### Plastic Photon

So 1/x is no longer rational?

15. Feb 23, 2007

### gammamcc

is too

(no square root in that one)

16. Feb 23, 2007

### Plastic Photon

what does the square root matter?

17. Feb 23, 2007

### Plastic Photon

In that case, u=2x-1

1/u(((u+1)2)^2-(u+1)/2)^1/2

18. Feb 23, 2007

### Hootenanny

Staff Emeritus
Look up the definition of a polynomial.

19. Feb 23, 2007

### rasensuriken

Thanks vietdao29. Ya i can go from there...I am a form 4 student studying integration on my own. Hope these can help me in form 5 and form 6.

20. Feb 24, 2007

### gammamcc

There is an easier way. I had the quickest way expect that I got diss'ed over partial fractions. You shouldn't have to rely on megaformulas.