please show me the steps to integrate 12x/(3x^2-2)?
回答 (3)
2017-02-21 7:17 pm
Let u = 3x² - 2
Then, du = 6x dx
∫ [12x / (3x² - 2) dx
= ∫ [2/(3x² - 2)] * ( 6x dx)
= ∫ (2/u) du
= 2 ln|u| + C
= 2 ln|3x² - 2| + C
Then, du = 6x dx
∫ [12x / (3x² - 2) dx
= ∫ [2/(3x² - 2)] * ( 6x dx)
= ∫ (2/u) du
= 2 ln|u| + C
= 2 ln|3x² - 2| + C
2017-02-21 6:37 pm
Use substitution.
Let u = 3x²–2
du = 6x dx
∫(2/u)du
=2 ln(u) + c
=2 ln(3x²–2) + c
Let u = 3x²–2
du = 6x dx
∫(2/u)du
=2 ln(u) + c
=2 ln(3x²–2) + c
2017-02-21 6:49 pm
You know that the integral of (1/x) dx is ln(x)
Also, d/dx[ln[f(x)] is f (x)/f(x)
So one approach is to differentiate kln(3x^2 - 2) which is 6kx/(3x^2 - 2)
and from 6k = 12 work out that the integral should be 2ln(3x^2 - 2)
Also, d/dx[ln[f(x)] is f (x)/f(x)
So one approach is to differentiate kln(3x^2 - 2) which is 6kx/(3x^2 - 2)
and from 6k = 12 work out that the integral should be 2ln(3x^2 - 2)
收錄日期: 2021-05-01 21:32:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170221102656AAV4ThQ