Posted by Erika on Friday, May 24, 2013 at 12:52am.
Can you check this please
i'm not sure why it's wrong .__.
1) indefinite integral of 1/t *ln(t) dt
the t*ln(t) is altogether on the bottom
i got ln(ln(x)) + C
2) find the indefinite integral of 1/(sqrt(t)*[1-2*sqrt(t)]dt
sqrt(x) is on the bottom.
i got -ln(1-2*sqrt(t))+C
can you tell me why mine's answers are wrong? thnx
- Integrals- Log/Ln - Kuai, Friday, May 24, 2013 at 1:28am
1. u = ln(t)
du = 1/t dt
u^2/2 + c
(ln(2))^2/2 + c
2. 1/t^2 (1- 2t^2)dt
Then multiply 1/t^2
-1/t -2t + c
- Integrals- Log/Ln - Kuai, Friday, May 24, 2013 at 1:31am
(ln(t))^2/2 + c
- Integrals- Log/Ln - Erika, Friday, May 24, 2013 at 1:38am
I'm still getting it wrong, so I'm not sure what's with that.. Thank you for your help though!
- Integrals- Log/Ln - Steve, Friday, May 24, 2013 at 11:02am
#1 is correct: ln(lnx))+C
#2: If you mean
∫ 1/√t * (1-2√t) dt
Let u = 1-2√t
du = -1/√t dt
and you have
= -1/2 u^2 + C
= -1/2 (1-2√t)^2 + C
= -1/2 (1 - 4√t + 4t) + C
= -2 + 2√t - 2t + C
you can absorb the -2 into C, leaving
2√t - 2t + C
- Integrals- Log/Ln - Steve, Friday, May 24, 2013 at 11:07am
Or, if you mean
∫ 1/(√t * (1-2√t)) dt
Again let u=1-2√t
du = -1/√t dt
and we have
∫ -1/u du
= -lnu + C
= -ln(1-2√t) + C
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