# Integrals- Log/Ln

posted by on .

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 - ,

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 - ,

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

• Integrals- Log/Ln - ,

I'm still getting it wrong, so I'm not sure what's with that.. Thank you for your help though!

• Integrals- Log/Ln - ,

#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 - ,

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