I wish to verify if my answer is correct:
Find $\int_C \vec{F}\cdot\,d\vec{R}$ where $\vec{F}(x,y) = \langle \tanh^2x, -3x\rangle$ and $C$ is the portion of the parabola $x = (y+1)^2$ from (1,0) to (4,1).
My solution: Parametrize the parabola $x = (y+1)^2$ as $\vec{R}(t) = \langle t, |\sqrt{t}|-1\rangle$. Since $x$ ranges from 1 to 4, then $1 \leq t \leq 4$ as well. Now,
\begin{align*} \text{Let }\vec{F} &= \langle \tanh^2x, -3x\rangle\\ \vec{F}(\vec{R}(t)) &= \langle \tanh^2t, -3t\rangle \end{align*} and \begin{align*} \vec{R}'(t) &= \left\langle 1, \dfrac{1}{2\sqrt{t}} \right\rangle. \end{align*} Hence, \begin{align*} \vec{F}(\vec{R}(t))\cdot\vec{R}'(t) &= \langle \tanh^2t, -3t\rangle \cdot \left\langle 1, \dfrac{1}{2\sqrt{t}} \right\rangle = \tanh^2t - \dfrac{3\sqrt{t}}{2} \end{align*} Thus, \begin{align*} \int_{C_2} \vec{F}(\vec{R}(t))\cdot\vec{R}'(t)\,dt &= \int_1^4 \tanh^2t - \dfrac{3\sqrt{t}}{2}\,dt\\ &= \ln(e^{-2t}+1) + \dfrac{2e^{-2t}}{2e^{-2t} + e^{-4t} + 1} - t^{\frac{3}{2}}\bigg|_1^4\\ &= \fbox{$\ln(e^{-8}+1) + \dfrac{2e^{-8}}{2e^{-8} + e^{-16} + 1} - 8 - \left(\ln(e^{-2}+1) + \dfrac{2e^{-2}}{2e^{-2} + e^{-4} + 1} - 1\right)$} \end{align*}
My answer really looks awkward, which is why I am asking if it is correct. Otherwise, where did I go wrong?
Something seems to have gone wrong with your integral. Note that, $\displaystyle \int \tanh^2 (t) = t - \tanh (t) + C$
As $ \displaystyle \tanh(t) = \frac{e^{2t} - 1}{e^{2t}+1}$ ,
$\displaystyle I = \int \tanh^2 (t) = \int \left(\frac{e^{2t} -1}{e^{2t}+1}\right)^2 \ dt = \int \frac{(e^{2t} + 1)^2 - 4e^{2t}}{(e^{2t}+1)^2} \ dt$
$ \displaystyle = \int \: dt \: - 2 \int \frac{2e^{2t}}{(e^{2t} + 1)^2} \ dt$
In second integral, substituting $e^{2t}+1 = u, \ 2 e^{2t} dt = du$ and that leads to, $ \displaystyle I = t + \frac{2}{e^{2t}+1} + C$
It can also be written as $ \displaystyle I = t - \left(1 - \frac{2}{e^{2t}+1}\right) + C_1 = t - \tanh(t) + C_1$
So the line integral should be $\displaystyle \left[t + \frac{2}{e^{2t}+1} - t^{3/2} \right]_1^4 = \frac{2}{1+e^8} - \frac{2}{1+e^2} - 4$
or $ \ \displaystyle \tanh(1) - \tanh(4) - 4$