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August 2, 2014

August 2, 2014

Posted by **allison** on Wednesday, October 8, 2008 at 6:38pm.

and integrate

x/x^2 dx

it might be simple...i just need a head start...tnx

- integrate -
**allison**, Wednesday, October 8, 2008 at 6:39pmdo i use the subsitution rule?

- integrate -
**allison**, Wednesday, October 8, 2008 at 6:44pmto be more specific...i know its best for me to attempt the problem

so for the first problem i used substitution as

v=3u

dv=3du

du=dv/3

so answer i got was 1/3 ln3u

is that correct?

and on the second on i tried

x*x^-2 as x^-3 dx so

ans is -1/2x^-2

is that correct? please let me know before i go ahead...i have a test tommorow and i am trying to do as many practice problem as i can...tnx

- integrate -
- integrate -
**Damon**, Wednesday, October 8, 2008 at 8:21pmdu/3u = (1/3) du/u

which is (1/3)[ ln u + ln constant ]

--> (1/3) ln ( c u)

c = 3 works

so

(1/3) ln 3u + constant

x/x^2 = 1/ x

so

dx/x --> ln x + constant

I think x^1*x^-2 = x^-1 = 1/x

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