Could you please just check if I solved this integral correctly?
$\int_\gamma ((\frac{2x}{x^2+y^2}+\cos x)dx+\frac{2(x^2+y^2+y)}{x^2+y^2}dy)$
where $\gamma = \{x(t)=e^{2t}\cos t, y(t)=te^{3t^2}\}$
Firstly I reordered the integral like (1) + (2):
$\int_\gamma (\frac{2x}{x^2+y^2}dx+\frac{2y}{x^2+y^2}dy)+\int_\gamma (\cos x dx + 2dy)$
Then solved (2):
\int_0^1 (\cos(x(t))x'(t)+2y'(t))dt=[\sin(e^{2t}\cos t)+2te^{3t^2}]_0^1= $=\sin(e^2 \cos 1)+2e^3-\sin 1$
Then I verified that (1) is a conservative field ($\int (Mdx+Ndy)$): $M_y=\frac{4xy}{(x^2+y^2)^2}=N_x$
and looked for the potential function $F(x,y)$:
$F(x,y)=\int Mdx = \int \frac{2x}{x^2+y^2}=\ln(x^2+y^2)+c(y)$ then found c'(y) with $F_y=N $
\frac{2y}{x^2+y^2}+c'(y)=\frac{2y}{x^2+y^2}
so c'(y) = 0 and $c(y)$ = k and $F(x,y)=\ln(x^2+y^2)+k$.
By path independence I solved (2) as:
$F(x(1),y(1))-F(x(0),y(0))=\ln(e^4\cos(1)^2)+e^6)=\ln(\cos(1)^2+e^2)+4$
So in total (1)+(2) is:
$\ln(\cos(1)^2+e^2)+4+\sin(e^2 \cos 1)+2e^3-\sin 1$
It took me ages to type this in, so I hope it's a "good" question.