Partial fraction decomposition

[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?

 

[konthelion]

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

 

[rock.freak667]

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

 

[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.

 

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

 

[rock.freak667]

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.

 

[Ry122]

The integral of 1/x is lnx + c
So then shouldn't 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.

 

[konthelion]

Integrals of the form a+bu

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

So then shouldn't 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.

 

[rock.freak667]

The integral of 1/x is lnx + c
So then shouldn't 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

 

[Ry122]

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

 

[Dick]

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

 

[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.

 

[Defennder]

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

 

[Ry122]

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

 

[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?

 

[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

 

[m_s_a]

http://www.imagehuge.com/images/a7ttelgdzf3crvdsouk.jpg