Math Help: Integration by parts

[SlowToady]

Thought I'd throw this out to the PPOT crowd.

Evaluate the integral from 1 to 4 of sqrt(t)*ln(t)dt

For the parts I set

u = ln(t)
du = (1/t)dt

dv = sqrt(x)
v = (1/(2sqrt(t)))dt

Using the integration by parts formula I get:

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

I was thinking of doing I.b.P again, but I don't see it eliminating either of the terms allowing me to integrate.

Anyone have any ideas?

[M.D. Holloway]

int{x*sqrt(1-x^2)dx}

let u=1-x^2.....(1)

then,du/dx= -2x giving dx= -du/2x

int{x*sqrt(1-x^2)dx}
=int{x*sqrt(u)*(-du/2x)}
= (-1/2)*int(sqrt(u))du
= -(1/2)*u^(3/2)/(3/2) +C
= -(1/2)*(2/3)*u^(3/2)+C
= -(1/3)*(u)^(3/2)+C
but,from (1), u=1-x^2

substituting for u,

int{x*sqrt(1-x^2)dx} = -(1/3)*(1-x^2)^(3/2)

[YTNUKLR]

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...