Arc Length Problem I am currently in the middle of the following problem.

Reparametrize the curve $\vec{\gamma } :\Bbb{R} \to \Bbb{R}^{2}$ defined by $\vec{\gamma}(t)=(t^{3}+1,t^{2}-1)$ with respect to arc length measured from $(1,-1)$ in the direction of increasing $t$.

By reparametrizing the curve, does this mean I should write the equation in cartesian form? If so, I carry on as follows.
$x=t^{3}+1$ and $y=t^{2}-1$
Solving for $t$
$$t=\sqrt[3]{x-1}$$
Thus,
$$y=(x-1)^{2/3}-1$$
Letting $y=f(x)$, the arclength can be found using the formula
$$s=\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}\cdot dx$$
Finding the derivative yields
$$f'(x)=\frac{2}{3\sqrt[3]{x-1}}$$
and 
$$[f'(x)]^{2}=\frac{4}{9(x-1)^{2/3}}.$$
Putting this into the arclength formula, and using the proper limits of integration (found by using $t=1,-1$ with $x=t^{3}+1$) yields
$$s=\int_{0}^{2}\sqrt{1+\frac{4}{9(x-1)^{2/3}}}\cdot dx$$
I am now unable to continue with the integration as it has me stumped. I cannot factor anything etc. Is there some general way to approach problems of this kind?

Hint:
Substitute $(x-1)^{1/3}=t$. Your integral will boil down to $$\int_{-1}^1t\sqrt{4+9t^2}\rm dt$$
Now set $4+9t^2=u$ and  note that $\rm du=18t~~\rm dt$ which will complete the computation. (Note that you need to change the limits of integration while integrationg over $u$.)

A Longer way:
Now integrate by parts with $u=t$ and $\rm d v=\sqrt{4+9t^2}\rm dt$ and to get $v$, you'd like to keep $t=\dfrac{2\tan \theta}{3}$
