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?

Your integration could be done.  However, there is a much easier way.  Calculate the arc-length using the parametric version of the curve.  
We have $x=u^3+1$ and $y=u^2-1$. (I changed the names of the parameters because I want to reserve $t$ for the parameter of the endpoint.) Then $\frac{dx}{du}=3u^2$ and $\frac{dy}{du}=2u$. Thus the arclength from $u=0$ to $u=t$ is given by
$$\int_0^t \sqrt{\left(\frac{dx}{du}\right)^2 +\left(\frac{dy}{du}\right)^2}\,du.$$
We have used the parametric arclength formula, much easier!  The integration starts at $u=0$, since that is the value of the parameter that gives us the point $(1,-1)$.
We end up integrating $\sqrt{9u^4+4u^2}$. Since $u\ge 0$, we can replace this by $u\sqrt{9u^2+4}$. Integrate, making the substitution $w=9u^2+4$. We arrive at
$$\frac{1}{27}\left((9t^2+4)^{3/2}-8\right).\qquad (\ast)$$
This is the arclength $s$, expressed as a function of $t$.
We want to parametrize in terms of $s$.  So solve for $t$ in terms of $s$, using $(\ast)$.  When you solve, there will be two candidate values of $t$. Take the non-negative one, since we started at $t=0$ and were told that t$ is increasing.  
Finally, in the original parametrization, replace $t$ by its value in terms of $s$.   
