# Line integral and green's theorem

I want to evaluate this line integral $$\int \limits_{L} \frac{2x(2-e^y)}{(1+x^2)^2} dx+ (3x +\frac{e^y}{1+x^2}) dy$$ where $$L$$ is the line of the unit circle $$x^2+y^2=1$$ from $$(0,1)$$ to $$(1,0)$$

My attempt : i used green's theorem to get to $$\int \int \limits_{D} 3 dxdy$$

And the area of a unit circle is $$\pi$$ so one forth is $$\frac{\pi}{4}$$

We are going clockwise on the line so the final answer is negative and i must multiply by $$3$$ to get $$-\frac{3\pi}{4}$$,

The problem is that the correct answer is $$e-\frac{3}{2} -\frac{3 \pi}{4}$$.

What is the mistake in my solution? and how to solve it correctly ?

The issue here is that the curve $$L$$ doesn't bound a region (and thus Green's theorem doesn't apply). You have assumed that it bounds the quarter of the unit disk lying in the first quadrant (what you are calling $$D$$), when in fact you need two more line segments to bound this region (the line segment from (1,0) to (0,0) and the line segment from (0,0) to (0,1)). In order to use Green's theorem you should define a new curve $$L'$$ consisting of all three of these pieces, use Green's theorem to write the integral over $$L'$$ as an integral over $$D$$, perform the integration, and then subtract the integrals over the two line segments.
When you use Green's theorem, you're also counting the line integral from $$(1,0)$$ to $$(0,0)$$ to $$(0,1)$$, so you need to subtract those off. Fortunately, the parameterizations of those two line segments make the integrals pretty easy. Be sure and keep the clockwise orientation going.