# Homework Help: 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:

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

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.