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

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

Typo

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

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

#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

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

Sure, I'd be happy to help you with these integrals. Let's go through them one by one.

1) Indefinite integral of 1/t * ln(t) dt:

To solve this integral, we can use the technique of integration by parts. The general formula for integration by parts is:

∫ u * v dx = u * ∫ v dx - ∫ (u' * ∫ v dx) dx

In this case, we can let u = ln(t) and dv = 1/t dt. Taking the derivatives and integrals, we have du = (1/t) dt and v = ln(t). Applying the integration by parts formula, we get:

∫ (1/t * ln(t)) dt = ln(t) * ln(t) - ∫ (ln(t) * (1/t) dt)

Now, let's focus on the remaining integral. We can simplify it further:

∫ (ln(t) * (1/t) dt) = ∫ (ln(t) / t) dt

To solve this integral, we can use substitution. Let's substitute u = ln(t). Taking the derivatives, we have du = (1/t) dt. Rewriting the integral in terms of u, we get:

∫ (ln(t) / t) dt = ∫ u du

Integrating u with respect to u gives u^2/2 + C. Substituting back u = ln(t), we obtain:

∫ (ln(t) / t) dt = (ln(t))^2/2 + C

Therefore, the correct answer is ((ln(t))^2)/2 + C, not ln(ln(t)) + C.

2) Indefinite integral of 1/(sqrt(t) * (1 - 2 * sqrt(t))) dt:

To solve this integral, we can use a substitution. Let's substitute u = sqrt(t). Taking the derivatives, we have du = (1/2) * (1/sqrt(t)) dt. Rearranging the equation, we get dt = 2 * u * du. Substituting u and dt in the integral, we have:

∫ (1/(sqrt(t) * (1 - 2 * sqrt(t))) dt = ∫ (1/(u * (1 - 2u))) * (2u du)

Simplifying further, we get:

∫ (1/(sqrt(t) * (1 - 2 * sqrt(t))) dt = ∫ (2 du / (1 - 2u))

Now, we can solve this integral directly:

∫ (2 du / (1 - 2u)) = -2ln|1 - 2u| + C

Substituting back u = sqrt(t), we have:

∫ (1/(sqrt(t) * (1 - 2 * sqrt(t))) dt = -2ln|1 - 2sqrt(t)| + C

Therefore, the correct answer is -2ln|1 - 2sqrt(t)| + C, not -ln(1 - 2 * sqrt(t)) + C.

I hope this clarifies why your answers were incorrect. Feel free to ask if you have any further questions!