- #1

- 565

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Attempt:

A/(2x-3)

Compare coefficients A = 2

[tex] 2 \int (1/(2x-3)) = 2ln(2x-3)[/tex]

This is wrong. Can someone please tell me what I'm doing wrong?

- Thread starter Ry122
- Start date

- #1

- 565

- 2

Attempt:

A/(2x-3)

Compare coefficients A = 2

[tex] 2 \int (1/(2x-3)) = 2ln(2x-3)[/tex]

This is wrong. Can someone please tell me what I'm doing wrong?

- #2

- 238

- 0

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

- #3

rock.freak667

Homework Helper

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you don't need partial fractions for this.

Attempt:

A/(2x-3)

Compare coefficients A = 2

[tex] 2 \int (1/(2x-3)) = 2ln(2x-3)[/tex]

This is wrong. Can someone please tell me what I'm doing wrong?

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

- #4

- 565

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- #5

- 238

- 0

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, [tex]\int \frac{1}{2x-3} \neq ln(2x-3)+C[/tex] therefore [tex]2 \int \frac{1}{2x-3} \neq 2ln(2x-3)+C[/tex]

Also, [tex]\int \frac{1}{2x-3} \neq ln(2x-3)+C[/tex] therefore [tex]2 \int \frac{1}{2x-3} \neq 2ln(2x-3)+C[/tex]

Last edited:

- #6

rock.freak667

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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

- 565

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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

- 238

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Integrals of the form a+bu

[tex]\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+C[/tex]

[tex]\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+C[/tex]

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

1/2x-3 be ln(2x-3) + c

Last edited:

- #9

rock.freak667

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[tex]\int \frac{1}{2x-3} dx[/tex]

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.

if u=2x-3, then [itex]\frac{du}{dx}=2 \Rightarrow \frac{du}{2}=dx[/itex]

now just put that into the integral

- #10

- 565

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

- #11

Dick

Science Advisor

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- #12

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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

Defennder

Homework Helper

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- #14

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

- #15

Gib Z

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Well, see if you can follow this:

[tex] \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)[/tex].

How about now?

- #16

- 565

- 2

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

Edit: nevermind, figured it out.

THanks for your help everyone

Last edited:

- #17

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

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