Tricky to 2 variable line/path integral

[Unart]

Homework Statement

Compute ∫f ds for [tex]f(x,y)=\frac {y^3}{x^7}[/tex] [tex]y=\frac {x^4}{4}[/tex] for [itex]1≤x≤2[/itex]

Homework Equations

∫f ds= ∫f(c(t))||c'(t)||

||c'(t)|| is the magnitude of ∇c'(t)

The Attempt at a Solution

From this... I gathered the following, by saying x=t
[tex]c(t)= <t,\frac {t^4}{4}>[/tex]
[itex]c'(t)= <1, t^3>[/itex]

That gave me a magnitude of [tex]\sqrt{1+t^6}[/tex]

so now I get that ∫f ds... through plugging in the magnitude and c(t) in the integral... this

[itex]∫\frac{t^5 \sqrt{1+t^6}}{64}[/itex] where [itex] x=t\, and\, 1≤t≤2 [/itex]

This is the part where I'm particularly stuck... am I missing a step or did I make a math error? How do I integrate this?

 

[LCKurtz]

Let ##u = 1+t^6##.

 

[Unart]

I am completely lost... my anti-derivative and integration is shaky at best.
so
u= 1+t^6
so do I just take the anti-derivative of t^5... and get t^6/(6*64) and then take the and then the anti-derivative of u which would [tex]\frac{2u^\frac{3}{2}}{3} [/tex] then anti-derivative of U which would t+t^7/7

 

[Unart]

no wait, there would be a product rule... too wouldn't it.

or am I wrong

 

[LCKurtz]

Ignoring the ##\frac 1 {64}##, you have$$
\int t^5\sqrt{1+t^6}\, dt$$If you let ##u = 1+t^6## then ##du = 6t^5dt##. That gives you$$
\frac 1 6\int u^{\frac 1 2}\, du$$You integrate that and substitute ##u = 1+t^6## in your answer. Then put in your limits.