SlowToady-here's how I did it.....
u=ln(t)
du=1/t dt
v=2/3t^(3/2)
dv=sqrt(t)dt
Think of dv as "t^(1/2)" rather than a sq. rt. Then you will get v correct, this is your problem.
Do I.B.P. again.....
integral [ln (t)(sqrt(t)]=ln(t)*[2/3t^(3/2)] - integral {[2/3t^(3/2)][1/t]}dt (remember to include 1 & 4 to solve for a definite answer)
answer=4.28245
Let me know if this makes sense/doesn't...
