Fin antiderivative $F(t)$ of $f(t)=(\ln x)^2$ such that $F(1)=0$

Question

I have this question:

find the antiderivative $F\left(t\right)$ of $f\left(t\right)=ln^2x$ such that $F\left(1\right)=0$

I just don't understand very well how to do this if they are asking $F\left(t\right)$ and $f\left(t\right)$ is a function of x

I did something and with substitution and integration by parts I got:

$F\left(t\right)=t\cdot ln^2t-2t\cdot lnt+2t-2$

but originally the 't' was 'x'. I just changed it. Is my answer correct? If not I would appreciate some help in understanding these type of questions :)

Thank you!

Answer 1

Yes, your answer is correct. For others that want to see how he got it: $$ F(t) = \int\ln^2{t}dt \\ u=\ln{t} \\ t=e^u \\ du=\frac{dt}{t} \\ F(t) = \int{u^2tdu} \\ = \int{u^2e^udu} \\ =e^u u^2 - 2\int{ue^u du} \\ =e^u u^2 - 2(ue^u - \int{e^u du}) \\ =e^u u^2 - 2ue^u + 2\int{e^u du} \\ F(t)=e^u u^2 - 2ue^u + 2e^u \\ F(t) = t\ln^2{t} - 2t\ln{t} + 2t + c \\ F(1) = 0 \\ 0 = \ln^2{1} - 2\ln{1} + 2 + c \\ 0 = 0 - 0 + 2 + c \\ c = -2 \\ F(t) = t\ln^2{t} - 2t\ln{t} + 2t - 2 $$

Answer 2

You had it right. Integrate by parts twice,

$$F(x)= F(1) + \int_1^x \ln^2tdt = x\ln^2 x - 2\int_1^x \ln t dt = x\ln^2 x -2x\ln x +2(x-1)$$

Answer 3

If the question is right then it can be a trick question

If x dosent depend on t :

Integral of (ln x)^2 with respect to t will be $t (ln x)^2 + c$

Since F(1) = 0 ...... $F(t) = (t-1)(ln x)^2$

If x depends on t....it will be ∫ [(ln x)^2 g'(x) dx] where t = g(x)

So if the question is absoultely complete then your answer is correct only for when x = t