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?
 A: Here we apply a substitution to get rid of the square root. Nevertheless the evaluation of the integral is somewhat cumbersome. We have a function
\begin{align*}
&f:\mathbb{R^2}\to\mathbb{R}\\
&f(x,y)=x-y
\end{align*}
and a curve
\begin{align*}
C=\{(x,y):y=2(x+1)^2,-1\leq x\leq 0\}
\end{align*}
where we want to evaluate the line integral clockwise along $C$ from the point $(0,2)$ to $(-1,0)$. The parametrisation in this case is
\begin{align*}
x&=x(t)\\
y&=y(t)=2(x(t)+1)^2\\
r^{\prime}(t)&=\left(x^{\prime}(t),4\left(x(t)+1\right)x^{\prime}(t)\right)\\
\|r^{\prime}(t)\|&=\left|x^{\prime}(t)\right|\,\sqrt{1+16\left(x(t)+1\right)^2}\tag{1}
\end{align*}
Looking at (1) indicates the substitution
\begin{align*}
\color{blue}{x}&=\color{blue}{\frac{1}{4}\sinh(t)-1}\\
y&=\frac{1}{8}\sinh^2(t)\\
\|r^{\prime}(t)\|&=\frac{1}{4}\left|\cosh(t)\right|\,\sqrt{1+\sinh^2(t)}\\
&=\frac{1}{4}\cosh^2(t)
\end{align*}
The variable change transforms the interval of integration into
\begin{align*}
-1\leq &x\leq 0\\
0\leq &t\leq \sinh^{-1}(4)
\end{align*}

We so obtain with some help of Wolfram Alpha
\begin{align*}
&\color{blue}{\int_{\sinh^{-1}(4)}^0\left(-\frac{1}{8}\sinh^2(t)+\frac{1}{4}\sinh(t)-1\right)\frac{1}{4}\cosh^2(t)dt}\\
&\qquad=\frac{1}{32}\int_{0}^{\sinh^{-1}(4)}\left(\sinh^4(t)-2\sinh^3(t)+9\sinh^2(t)-2\sinh(t)+8\right)dt\\
&\qquad\,\,=\frac{1}{768}\left(16+508\sqrt{17}+93\sinh^{-1}(4)\right)\color{blue}{\simeq 3.001\,8}
\end{align*}

A: 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}
