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March 30, 2015

March 30, 2015

Posted by **Erika** on Friday, May 24, 2013 at 12:52am.

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 1:28am1. 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 1:31amTypo

(ln(t))^2/2 + c

- Integrals- Log/Ln -
**Erika**, Friday, May 24, 2013 at 1:38amI'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

∫-u du

= -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 11:07amOr, 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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