I'm struggling to compute arc length (multivariable calculus)

[Physics_Is_Fun]

Homework Statement

Find the arc length of the curve
(t) = (1; 3t2; t3) over the interval 0  t  1.

Homework Equations

L=sqrt(f'(t)^2+g'(t)^2+...+n'(t)^2) (integrated from a to b)
int(udv)=uv-int(vdu)

The Attempt at a Solution

Seems like it should be fairly straightforward-- the derivative vector ends up being (0, 6t, 3t^2)
1.) L=int[sqrt(0^2+(6t)^2+(3t^2)^2)] (over 0 to 1)
2.) L=int[sqrt(36t^2+9t^4)] (from 0 to 1)
3.) L=3int[t*sqrt(t^2+4)] (from 0 to 1)
4.) From here, I wanted to solve using integration by parts. However, that seemed to only result in a more complicated integral no matter whether I set t as u or as dv.

Can anyone give me a hint? I'd like to work it out on my own, but an idea as to which method to use would be much appreciated.
Thanks so much to anyone who takes the time to respond to this. :)

 

[ehild]

Homework Statement

Find the arc length of the curve
(t) = (1; 3t2; t3) over the interval 0  t  1.

Homework Equations

L=sqrt(f'(t)^2+g'(t)^2+...+n'(t)^2) (integrated from a to b)
int(udv)=uv-int(vdu)

The Attempt at a Solution

Seems like it should be fairly straightforward-- the derivative vector ends up being (0, 6t, 3t^2)
1.) L=int[sqrt(0^2+(6t)^2+(3t^2)^2)] (over 0 to 1)
2.) L=int[sqrt(36t^2+9t^4)] (from 0 to 1)
3.) L=3int[t*sqrt(t^2+4)] (from 0 to 1)
4.) From here, I wanted to solve using integration by parts. However, that seemed to only result in a more complicated integral no matter whether I set t as u or as dv.

Can anyone give me a hint? I'd like to work it out on my own, but an idea as to which method to use would be much appreciated.
Thanks so much to anyone who takes the time to respond to this. :)

Us integration by substitution. Is not the factor t connected to the derivative of t²+4?

ehild

 

[Physics_Is_Fun]

Thank you!

 

[Physics_Is_Fun]

Can't believe I missed that...