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)*[12*sqrt(t)]dt
sqrt(x) is on the bottom.
i got ln(12*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 du
u^2/2 + c
(ln(2))^2/2 + c
2. 1/t^2 (1 2t^2)dt
Then multiply 1/t^2
(1/t^2 2)dt
1/t 2t + c

Integrals Log/Ln  Kuai, Friday, May 24, 2013 at 1:31am
Typo
(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 * (12√t) dt
Let u = 12√t
du = 1/√t dt
and you have
∫u du
= 1/2 u^2 + C
= 1/2 (12√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 * (12√t)) dt
Again let u=12√t
du = 1/√t dt
and we have
∫ 1/u du
= lnu + C
= ln(12√t) + C
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