Unsure how to solve this integral Let $$F\left(x,y\right)=xi+e^{y^2}j.$$ Evaluate $$\int _CF\cdot dr,$$ where $C$ is the curve $$r\left(t\right)=\frac{1}{t^2+1}i+\left(3+10t^2\cos\left(\frac{\pi t}{2}\right)\right)j,\quad0\le t\le 1.$$
I am having a bit of difficulty with this as I have already tried setting it up as $$\int _0^1F\left(r\left(t\right)\right)\cdot r'\left(t\right)dt.$$ But I get really large equations for the derivative of $r(t)$ which leads to math that I just can't do on paper. I think I can use greens theorem but I'm not really sure where to start.
 A: Using the following fact:

Theorem: Let $F=P\vec{i}+Q\vec{j}$ be a vector field over a region simply connected $\Omega\subseteq \mathbb{R}^{2}$. Suppose that $P,Q\in \mathcal{C}^{1}(\Omega)$ and $$\frac{\partial P}{\partial y}(x,y)=\frac{\partial Q}{\partial x}(x,y)$$
for all $(x,y)\in \Omega$. Therefore $F$ is a conservative vector field and thus there exists a potential function $f$ such that $\nabla f=F$.

Define $P(x,y)=x$ and $Q(x,y)=e^{y^{2}}$ and notice that $$\frac{\partial P}{\partial y}(x,y)=0=\frac{\partial Q}{\partial x}(x,y)$$
for all $(x,y)\in \mathbb{R}^{2}$ so $F(x,y)=P(x,y)\vec{i}+Q(x,y)\vec{j}$ is a conservative vector field by the theorem.
Thus, there exists a function $f$ such that $\nabla f=F$.
Since $f$ there exists and satisfies $\nabla f=F$ so
$$f_{x}(x,y)=x,\quad f_{y}(x,y)=e^{y^{2}}$$
then integrating $f_{x}(x,y)=x$ respect to $x$ we have $f(x,y)=\frac{x^{2}}{2}+g(y)$ and then differentiating respect to $y$ we have $f_{y}(x,y)=g'(y)$ but also $f_{y}(x,y)=e^{y^{2}}$ so $g'(y)=e^{y^{2}}$ and integrating respect to $y$ we have $g(y)=\int e^{y^{2}}\, {\rm d}y$. Thus,
$$f(x,y)=\frac{x^{2}}{2}+\int_{0}^{y} e^{s^{2}}\, {\rm d}s$$
Notice that indeed $\nabla f=F$.
Therefore, by the Fundamental Theorem of Line Integration
\begin{align*}\int_{C}F\cdot\, {\rm d}r&=f(r(1))-f(r(0))\\&=f\left(\frac{1}{2},3\right)-f\left(1,3\right)\\
&=\frac{\left(\frac{1}{2}\right)^{2}}{2}+\int_{0}^{3}e^{s^{2}}\, {\rm d}s-\frac{1^{2}}{2}-\int_{0}^{3}e^{s^{2}}\, {\rm d}s,\\ &=\frac{1}{2^{3}}-\frac{1}{2},\\&=\boxed{-\frac{3}{8}}\end{align*}
