• rasensuriken
In summary: Sorry. I was off by a minus sign. : ) That's what I get for jumping into the middle of a thread without reading the previous posts. : )So we're even. No hard feelings, OK? : )In summary, a user asks for help with a calculus problem involving integration and provides a screenshot of the problem. Other users offer various methods and approaches to solve the problem, including completing the square, partial fraction decomposition, and substitution. The discussion also includes a brief debate about the definition of a rational function and a minor error is corrected.

#### rasensuriken

b]1. Homework Statement [/b]

Calculus, integration

## Homework Equations

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

## The Attempt at a Solution

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

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

rasensuriken said:
b]1. Homework Statement [/b]

Calculus, integration

## Homework Equations

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

## The Attempt at a Solution

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

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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:
$$ax + b = \frac{1}{t}$$
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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? :)

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Forget what I had written here

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gammamcc said:
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.

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

Schrodinger's Dog said:
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..

arildno said:
$$\sqrt{4^{2}-4}=\sqrt{16-4}=\sqrt{12}=2=4-\sqrt{4}$$
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 :/

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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.

gammamcc said:
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

Plastic Photon said:
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.

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

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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: 189
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tim_lou said:
$$u=\sqrt{x^2-x}$$
$$du=\frac{(2x-1)dx}{2\sqrt{x^2-x}}$$
$$\int{\frac{dx}{(2x-1)\sqrt{x^2-x}}}=\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{4u+1}}$$

$$\int{\frac{2du}{(2x-1)^2}}=\int{\frac{2du}{(2u)^2+1}}$$ Very easy now, form of arctan integral.

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

We can't. Your attachment is "Pending Approval" by a Mentor to make sure your not a predator give us viruses :)

Bad algebra? Where did he forget what u was?

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

that is the solution

Gib Z said:
Bad algebra? Where did he forget what u was?

Because

$$u=/sqrt{x^2-x}$$

$${(2x-1)^2}={4x^2}-4x+1={4u^2}+1$$

oops, forgot the square...fixed. hehe

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OK PP. I needed more convincing.

Did you guys notice that I fixed tim_lou's error in the post right after his?

OK. If it comes down to pistols at dawn, I'll be busy.

: )