- #1

amcavoy

- 665

- 0

[tex]\int\left(\frac{1-x}{x^2-4}\right)dx[/tex]

Thanks.

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- Thread starter amcavoy
- Start date

- #1

amcavoy

- 665

- 0

[tex]\int\left(\frac{1-x}{x^2-4}\right)dx[/tex]

Thanks.

- #2

LeonhardEuler

Gold Member

- 860

- 1

[tex]\int-\frac{x-1}{x^2-4}dx[/tex]

[tex]=-\int\frac{x-2}{(x+2)(x-2)}+\frac{2}{2(x+2)(x-2)}dx[/tex]

[tex]=-\int\frac{1}{x+2}+\frac{2+x}{2(x+2)(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]

[tex]=-\int\frac{1}{x+2}+\frac{1}{2(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]

Now you can directly compute the first two parts as logs and the last as a log by recognizing x as one quarter the derivative of the denomenator

- #3

amcavoy

- 665

- 0

LeonhardEuler said:

[tex]\int-\frac{x-1}{x^2-4}dx[/tex]

[tex]=-\int\frac{x-2}{(x+2)(x-2)}+\frac{2}{2(x+2)(x-2)}dx[/tex]

[tex]=-\int\frac{1}{x+2}+\frac{2+x}{2(x+2)(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]

[tex]=-\int\frac{1}{x+2}+\frac{1}{2(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]

Now you can directly compute the first two parts as logs and the last as a log by recognizing x as one quarter the derivative of the denomenator

Well it looks like the only way to save time here is to use my TI-89 .

Thanks for the reply Euler.

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