- #1

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of 6/(5x-2)

--> I arrive at 6/5*ln(u), is this the right thing?

How would I evaluate something like ln(-2), or do I just assume it is divergent since it the other limit will come out to neg infinity?

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- Thread starter frasifrasi
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- #1

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of 6/(5x-2)

--> I arrive at 6/5*ln(u), is this the right thing?

How would I evaluate something like ln(-2), or do I just assume it is divergent since it the other limit will come out to neg infinity?

- #2

quasar987

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[tex]\int_{-\infty}^{0}\frac{6}{5x-2}dx = \frac{6}{5}\int_{-\infty}^{-2}\frac{du}{u}[/tex]

But obviously, it's not true that ln(u) is an antiderivative of 1/u on the domain (-\infinity,-2) simply because ln(u) is not defined there. So your change of variable was not so useful. Try u=2-5x.

- #3

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wait, do you mean 5x - 2?

- #4

Hurkyl

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- #5

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So, what substitution should I use?

- #6

Dick

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So, what substitution should I use?

Hurkyl is trying to get you to recall ln(|u|) is also an antiderivative for 1/u.

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I am still unsure wth I should do -- u-subs or what?

Thank you.

Thank you.

- #8

Dick

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I am still unsure wth I should do -- u-subs or what?

Thank you.

You already did the u-subs correctly. The antiderivative is (6/5)*ln(|5x-2|). Now you just have to think what happens if you put the limits in.

- #9

quasar987

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wait, do you mean 5x - 2?

No, I mean 2-5x. Try it.

- #10

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where did that come from, though?

- #11

quasar987

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But how was that derived, can anyone explain?

- #13

quasar987

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- #14

HallsofIvy

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If u is negative, |u|= -u. Since, on this range, 5x-2 is negative, |5x-2|= 2- 5x.

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