# Integrals by parts im stuck

1. Feb 1, 2006

### nick727kcin

u=lnt dv= t^.5 dt
du=1/t dt v= (2t^1.5)/3

i keep on getting stuck with this integral. any help would be appreciated

2. Feb 1, 2006

### nick727kcin

i dont want anyone to wast their time with this, this hw is due exaclty 6 hours from now, so if you dont see this thread until 5 hours after i originally posted it, dont even worry about it

3. Feb 1, 2006

### HallsofIvy

Staff Emeritus
It won't take long enough to worry about wasting time:
You already have $\int_1^4 \sqrt{t}(ln(t))dt$ and
you have set u= ln t so $du= \frac{1}{t}dt$ and $dv= t^{\frac{1}{2}}dt$ so that $v= \frac{2}{3}t^{\frac{3}{2}}$.

Okay, plugging that into the integration by parts formula gives
[tex]\int_1^4\sqrt{t}(ln(t))dt= \frac{2}{3}t^{\frac{3}{2}}ln(t)\left|_1^4- \frac{2}{3}\int_1^4 t^{\frac{1}{2}}dt[/itex]

Why would you have any difficulty with that?

4. Feb 1, 2006

### nick727kcin

oooooo i forgot to transfer the 2/3 to the left. thank you very much kind sir

:!!)