# Partial fraction decomposition

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?

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

rock.freak667
Homework Helper
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?

you don't need partial fractions for this.

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

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.

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

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rock.freak667
Homework Helper
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.

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.

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.

Integrals of the form a+bu

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

So then shouldnt the integral of
1/2x-3 be ln(2x-3) + 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.

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rock.freak667
Homework Helper
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.

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

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

now just put that into the integral

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

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

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.

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

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

Gib Z
Homework Helper
What, you mean how to divide?

Well, see if you can follow this:

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