Partial fraction decomposition

1. Jun 18, 2008

Ry122

Problem: Find the integral of 2dx/(2x-3) using partial fraction decomposition
Attempt:
A/(2x-3)
Compare coefficients A = 2
$$2 \int (1/(2x-3)) = 2ln(2x-3)$$
This is wrong. Can someone please tell me what I'm doing wrong?

2. Jun 18, 2008

konthelion

Use u-substitution. Let u=2x-3.

3. Jun 18, 2008

rock.freak667

you don't need partial fractions for this.

just put u=2x-3 and go from there

4. Jun 18, 2008

Ry122

This is just a simple problem that I want to apply the technique to and further my understanding of it so that I can begin using it on more difficult questions.

5. Jun 18, 2008

konthelion

Did the problem tell you that you must use partial fraction for this problem? Because in my opinion, it makes no sense to use it.

Also, $$\int \frac{1}{2x-3} \neq ln(2x-3)+C$$ therefore $$2 \int \frac{1}{2x-3} \neq 2ln(2x-3)+C$$

Last edited: Jun 18, 2008
6. Jun 18, 2008

rock.freak667

well in a sense...when you have a linear factor in the denominator like 2x-3, in the question, it kinda already is a partial fraction.

7. Jun 18, 2008

Ry122

The integral of 1/x is lnx + c
So then shouldnt the integral of
1/2x-3 be ln(2x-3) + c
If not then what is it?
And yes the question did ask me to use partial fractions for this problem.

8. Jun 18, 2008

konthelion

Integrals of the form a+bu

$$\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+C$$

To check that this is false: if you take the derivative with respect to x of "ln(2x-3) + c", it will not equal $$\frac{1}{2x-3}$$ since you have to use the chain rule.

Last edited: Jun 18, 2008
9. Jun 18, 2008

rock.freak667

$$\int \frac{1}{2x-3} dx$$

if u=2x-3, then $\frac{du}{dx}=2 \Rightarrow \frac{du}{2}=dx$

now just put that into the integral

10. Jun 18, 2008

Ry122

Can partial fraction decomposition even be used in this situation to get a correct answer?

11. Jun 18, 2008

Dick

The partial fraction decomposition of 1/(2x-3) IS 1/(2x-3). Please believe what people are telling you.

12. Jun 19, 2008

Ry122

How would i do this one:
integral of ((x^2)+1)/(6x-9x^2)
The highest degree of the denominator is equal to the highest degree of the numerator but it has to be greater than it.

13. Jun 19, 2008

Defennder

Do polynomial long division first, to make sure the degree of the denominator is greater than that of the denominator.

14. Jun 19, 2008

Ry122

Have no idea how to do that. Can you please show me?

15. Jun 19, 2008

Gib Z

What, you mean how to divide?

Well, see if you can follow this:

$$\frac{x^2 +1}{6x-9x^2} = \frac{-1}{9}\left( \frac{9x^2+9}{9x^2 - 6x} \right) = \frac{-1}{9}\left( \frac{9x^2 -6x + 6x +9}{9x^2 - 6x} \right) = \frac{-1}{9}\left( 1 + \frac{6x+9}{9x^2 - 6x} \right) = \frac{-1}{9}\left( 1+ \frac{2x+3}{3x^2 - 2x} \right)$$.

How about now?

16. Jun 19, 2008

Ry122

i understand but why did the sign of only the denominator change in the 1st step and not the numerator?
Edit: nevermind, figured it out.
THanks for your help everyone

Last edited: Jun 19, 2008
17. Jun 19, 2008

m_s_a

http://www.imagehuge.com/images/a7ttelgdzf3crvdsouk.jpg [Broken]

Last edited by a moderator: May 3, 2017
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