Evaluate the line integral along a parabola 
Evaluate the line integral of $f(x,y)=-y+x$ along part of the parabola $y=2(x+1)^2$ from the point $(0,2)$ to the point $(-1,0)$

I need help trying to find a good parameterization for this because what I've done just lands me in a mess.
My work so far:
Let $x=t, y=2(t+1)^2$
$$
\begin{split}
r(t)  &= \left<t,2(t+1)^2\right>, \quad -1\leq t \leq 0 \\
r'(t) &= \left<1,4(t+1)\right>,   \quad -1\leq t \leq 0 \\
\|r'(t)\| &= \sqrt{1+16(t+1)^2} \\
-y+x &= -2(t+1)^2+t\\
     &=-2t^2-3t-2\\
\end{split}
$$
So we have
$$
\int_0^{-1}\left(-2t^2-3t-2\right)\sqrt{1+16(t+1)^2}dt
$$

This integral is really ugly. so then I tried a different method:
$$
\begin{split}
y &= 2(x+1)^2 \\
\frac{dy}{dx} &= 4x+4 \\
dS &= \sqrt{1+(4x+4)^2} dx\\
   &=\sqrt{16x^2+32x+17} dx
\end{split}
$$
So we get
$$\int_0^{-1} \left(-2(x+1)^2+x\right)\sqrt{16x^2+32x+17}dx$$
Again, very ugly. Can someone please help me solve this?

Using the substitution $4(x+1)=y$ on the latter integrand, we get:
\begin{align}\frac{1}{16}\int_{0}^{1\over4}(\frac{y^2}{2}-y-4)\sqrt{y^2+1}\,dy\end{align}
The integral$\int_{0}^{1\over4}(y+4)\sqrt{y^2+1}\,dy$ is quite straightforwardly evaluated.
For the $4\int_{0}^{1\over4}\sqrt{y^2+1}\,dy$, substituting $y=\sinh u$, we get $2(3\ln(y+\sqrt{y^2+1})-y\sqrt{1+y^2})$.
For $\int_{0}^{1\over4}y\sqrt{y^2+1}\,dy$, let $y^2=\alpha$, getting $\frac{1}{3}(y^2+1)^{3\over2}$. The integral becomes:
\begin{align}\frac{1}{32}\biggr(\int_{0}^{1\over4}y^2\sqrt{y^2+1}\,dy\biggr)-\frac{1}{16}\biggr(6\ln(y+\sqrt{y^2+1})-2y\sqrt{1+y^2}+\frac{1}{3}(y+1)^{3\over2}\biggr)\biggr|_0^{1\over4}\end{align}
To integrate the first part, using the substitution $y=\sinh u$, it becomes:
\begin{align}\frac{1}{256}\biggr(2y^{3}\sqrt{1+y^2}+\ln(y+\sqrt{y^2+1})-y\sqrt{1+y^2}\biggr)\end{align}
So the integral finally is:
\begin{align}\frac{1}{256}\biggr(2y^{3}\sqrt{1+y^2}-95\ln(y+\sqrt{y^2+1})+31y\sqrt{1+y^2}-\frac{16}{3}(y+1)^{3\over2}\biggr)\biggr|_0^{1\over4}\end{align}
