Integrating a Line Integral over a Parabola: A Challenging Task?

[kbaumen]

Homework Statement

$$\int_{C}(xy+\ln{x})\mathrm{d}s$$

where C is the arc of the parabola $y=x^2$ from (1,1) to (3,9)

Homework Equations

$$\int_{C}f(x,y)\mathrm{d}s = \int_{a}^{b}f(x(t),y(t))\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\mathrm{d}t$$

The Attempt at a Solution

Ok, so I take x as the parameter t, so the parametric equations of the line $x=t$ and $y=t^2$ where $t\in [1,3]$. I can then rewrite this line integral as a regular definite integral by using the formula mentioned above (pardon the bad formatting). However, when I do that, I come up with

$$\int_{1}^{3}(t^3+\ln{t})\sqrt{1+4t^2}\mathrm{d}t$$

I can't seem to find a way to integrate this function. It is integrable, at least according to the Wolfram Integrator. A solutions manual for the book I'm studying from comes up with

$$\int_{1}^{3}(t^3+\ln{t})\sqrt{4t^2}\mathrm{d}t = \int_{1}^{3}(t^3+\ln{t})2t\mathrm{d}t$$

which is quite straightforward, however, it seems to me that it's a mistake since $\frac{dx}{dt}=1$ thus there is a 1 missing under the square root.

So can someone please help me find a way to integrate the integral I've come up with or am I missing something simple, i.e. it has been set up correctly in the book?

 

[Dickfore]

The solution manual is mistaken. The indefinite integral is not expressible in terms of elementary functions.

http://www.wolframalpha.com/input/?i=Integrate[(t^3+%2B+Log[t])+Sqrt[1+%2B+4+t^2],t]

 

[kbaumen]

Oh yeah, I had incorrectly input it in the wolfram integrator. The integral indeed cannot be expressed in terms of elementary functions.

So how do I go about the integral I have?

 

[Dickfore]

Wolfram Alpha gives an approximate numerical answer. I don't know if your integral is expressible in terms of some well known mathematical constants.

http://www.wolframalpha.com/input/?i=Integrate[(t^3+%2b+Log[t])+Sqrt[1+%2b+4+t^2],{t,1,3}]&incTime=true

 

[kbaumen]

Ok, cheers man.