I can't figure out this Intergral Someone help me Please....?

Question

The intergral from 1 to 4 of sqrt (t) times ln (t) dt
I tryed to use parts with U = sqrt t and dv = Ln(t) abd v=1/t but i can't figure out how to derive sqrt t besides 1/2 times 1/sqrt(t)

Answer 1

you should take u to be ln t and dv = sqrt t dt.

You get 2/3 t^3/2 ln t - 2/3 integral (sqrt t) dt.
you should know how to continue from here.

Answer 2

I'll do part of ∫ √t ln t dt and let you finish the job.
You have to use parts. Let
u = ln t dv = √t dt
du = dt/t v = 2/3 t^(3/2)
So ∫ √t ln t dt = 2/3 t^(3/2)* ln t - 2/3 ∫ √t dt,
= 2/3 t^(3/2)* ln t - 4/9 t^(3/2)
The way you tried it leads to a much harder problem!

Answer 3

Int sqrt(t)*ln(t) dt = 2/3 t^3/2*ln(t) -2/3 Int t^1/2dt=
2/3t^3/2 (ln t -2/3 ) taken between 1 to 4=
2/3*8(ln 4 -2/3)+4/9 =Check calculations

Answer 4

You better go the other way around.
Integral(t^(1/2)*ln(t)*dt)=
=2/3*t^(3/2)ln(t) - 2/3*Integral(t^(1/2)*dt)=
=2/3*t^(3/2)ln(t)-4/9*t^(3/2)
Now you can find your integral.

Answer 5

let u=ln(t) and dv=sqrt(t) dt.
then du=1/t dt, and v=(2/3)*(t^(3/2)).
Then integral becomes
uv-int(v du)
=(2*ln(t)/3)*(t^(3/2))-(2/3)*
int(sqrt(t) dt)
=(2*ln(t)/3)*(t^(3/2))-(4/9)*
t^(3/2)
=(2/3)*(t^1.5)*(ln(t)-(2/3))
Then plug your limits in and you're finished.