- #1

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1) x+2/(x^2-4) from 0 to 1

This just simplifies to 1/(x-2) so this should be ln(1-2)-ln(0-2)=ln(-1)-ln(-2)?

2) x/x+1, x/x^2+2...Do you use by parts? I cannot get any of these right where its of this nature..thanks so much

- Thread starter Jacob87411
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- #1

- 171

- 1

1) x+2/(x^2-4) from 0 to 1

This just simplifies to 1/(x-2) so this should be ln(1-2)-ln(0-2)=ln(-1)-ln(-2)?

2) x/x+1, x/x^2+2...Do you use by parts? I cannot get any of these right where its of this nature..thanks so much

- #2

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For number one, you integrated correctly. However, remember your logarithm rules.

[tex] ln(a) - ln(b) = ln(a/b) [/tex]

Here's a hint for two: If you have something such as[tex] \int \frac{x dx}{x+a} = \int \frac{x+a-a dx}{x+a} = \int 1dx - \int \frac{a dx}{x+a} [/tex]

You can generalize that further if you have the case x+ k on top. I think that may help.

Furthermore, on the x/x^2 + 2, try u-substitution.

[tex] ln(a) - ln(b) = ln(a/b) [/tex]

Here's a hint for two: If you have something such as[tex] \int \frac{x dx}{x+a} = \int \frac{x+a-a dx}{x+a} = \int 1dx - \int \frac{a dx}{x+a} [/tex]

You can generalize that further if you have the case x+ k on top. I think that may help.

Furthermore, on the x/x^2 + 2, try u-substitution.

Last edited:

- #3

VietDao29

Homework Helper

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This is wrong!!!Jacob87411 said:For some reason I just have a problem with these types of integrals..if someone could show me how to do one it would save me a lot of headaches..

1) x+2/(x^2-4) from 0 to 1

This just simplifies to 1/(x-2) so this should be ln(1-2)-ln(0-2)=ln(-1)-ln(-2)?

You are forgetting the absolute value. You should note that in the reals, ln(x) is

You can look again at the table of integrals, you may find something that reads:

[tex]\frac{dx}{x} = \ln |x| + C[/tex]

For the first one, as2) x/x+1, x/x^2+2...Do you use by parts? I cannot get any of these right where its of this nature..thanks so much

[tex]\int \frac{x}{x + 1} dx = \frac{x + 1 - 1}{x + 1} dx = \int dx - \int \frac{dx}{x + 1}[/tex].

For the second one, if your numerator is 1 degree less than your denominator, then it's common to try to take the derivative of the denominator, then rearrange the numerator so that it contains a multiple of the derivative of the denominator. Your answer will have a ln part and an arctan part (sometimes it does not have an arctan part).

[tex]\int \frac{x}{x ^ 2 + 2} dx[/tex]

Now:

(x

[tex]\int \frac{x}{x ^ 2 + 2} dx = \int \frac{\frac{1}{2} \times 2x}{x ^ 2 + 2}[/tex]

Now use the u-substitution:

u = x

[tex]\int \frac{x}{x ^ 2 + 2} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x ^ 2 + 2| + C = \frac{1}{2} \ln (x ^ 2 + 2) + C[/tex].

Since x

|x

Can you get it? :)

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