How do I integrate this, with respect to x?
1/(x^2-1)
Some step by step would be highly appriciated :) I have tried substitution, but cant get it to work since I get a ln(0) in my solution.
How do I integrate this?
2016-05-20 10:43 am
回答 (3)
2016-05-20 10:56 am
We can say 1/(x² - 1) => A/(x - 1) + B/(x + 1)
i.e. 1 = A(x + 1) + B(x - 1)
Letting x = 1 we have:
1 = 2A
so, A = 1/2
When x = -1 we have:
1 = -2B
so, B = -1/2
Hence, 1/(x² - 1) => 1/(2(x - 1)) - 1/(2(x + 1))
Then, ∫1/(x² - 1) dx => (1/2) ∫ [1/(x - 1) - 1/(x + 1)]
=> (1/2)[ln(x - 1) - ln(x + 1)]
i.e. (1/2)ln[(x - 1)/(x + 1)]
:)>
i.e. 1 = A(x + 1) + B(x - 1)
Letting x = 1 we have:
1 = 2A
so, A = 1/2
When x = -1 we have:
1 = -2B
so, B = -1/2
Hence, 1/(x² - 1) => 1/(2(x - 1)) - 1/(2(x + 1))
Then, ∫1/(x² - 1) dx => (1/2) ∫ [1/(x - 1) - 1/(x + 1)]
=> (1/2)[ln(x - 1) - ln(x + 1)]
i.e. (1/2)ln[(x - 1)/(x + 1)]
:)>
2016-05-20 10:51 am
1/(x^2 - 1) = 1/2 [(x+1)/(x^2 - 1) - (x-1)/(x^2 - 1)]
= 1/2 [1/(x-1) - 1/(x+1)]
integral = 1/2 [ln(x-1) - ln(x+1)]
= 1/2 [1/(x-1) - 1/(x+1)]
integral = 1/2 [ln(x-1) - ln(x+1)]
2016-05-20 10:48 am
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160520024358AAMfwl5