b]1. Homework Statement [/b]

Calculus, integration

## Homework Equations

http://img341.imageshack.us/img341/6091/questiontc3.jpg [Broken]

## The Attempt at a Solution

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

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
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:
Rewrite the argument x^2-x by "completing the square" . "u" substitution follows.

b]1. Homework Statement [/b]

Calculus, integration

## Homework Equations

http://img341.imageshack.us/img341/6091/questiontc3.jpg [Broken]

## The Attempt at a Solution

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

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

Last edited by a moderator:
VietDao29
Homework Helper
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:
Forget what I had written here

Last edited:
I vote for Man Monkey's approach.
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:
(x^2-x)^(1/2) = x - x^(1/2) is not true. You can check by multiplying the RHS with itself.

arildno
Homework Helper
Gold Member
Dearly Missed
What they're equivalent, just trying to help? Thought it might be easier if he got rid of the sqrt. $$\sqrt{4^{2}-4}=\sqrt{16-4}=\sqrt{12}=2=4-\sqrt{4}$$
Fascinating..

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

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.

I don't trust partial fractions since we don't have a rational function.
Your kidding right? So 1/((2x-1)(x^2-x)^1/2) is not a rational function?

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

So 1/x is no longer rational?

is too

(no square root in that one)

what does the square root matter?

In that case, u=2x-1

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

Hootenanny
Staff Emeritus
Gold Member
what does the square root matter?
Look up the definition of a polynomial.

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.

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.

$$u=\sqrt{x^2-x}$$
$$du=\frac{(2x-1)dx}{2\sqrt{x^2-x}}$$
$$\int{\frac{dx}{(2x-1)\sqrt{x^2-x}}}=\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{4u^2+1}}$$

Last edited:
I agree except for the last step.

In case no one was paying attention, see my attachment for the quick way.

#### Attachments

• 18 KB Views: 85
Last edited:
Gib Z
Homework Helper
$$u=\sqrt{x^2-x}$$
$$du=\frac{(2x-1)dx}{2\sqrt{x^2-x}}$$
$$\int{\frac{dx}{(2x-1)\sqrt{x^2-x}}}=\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{4u+1}}$$
$$\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{(2u)^2+1}}$$ Very easy now, form of arctan integral.

I don't agree. Forgetting what "u" is and applying bad algebra. Is anyone reading my attachment?