Line integral of $\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}-\frac{x}{(x+y)^2}\right)dy$ Determine the line integral $\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}-\frac{x}{(x+y)^2}\right)dy$ where $\gamma$ follows the circle from $(1,0)$ to $(0,1)$.
Let $x=\cos\theta$ and $y=\sin\theta$ where $0\leq\pi\leq\pi/2$. Then we have $\int_0^{\pi/2}\left(\left(\frac{1}{\cos\theta+1}-\frac{\sin\theta}{(\cos\theta+\sin\theta)^2}\right)(-\sin\theta)+\left(\frac{1}{\sin\theta+1}+\frac{\cos\theta}{(\cos\theta+\sin\theta)^2}\right)\cos\theta \right)d\theta$. Now putting this into symbolab I get $-0$ while the answer should be $1$.
But continuing  $\int_0^{\pi/2}\left(\frac{-\sin\theta}{\cos\theta+1}+\frac{\sin^2\theta}{(\cos\theta+\sin\theta)^2}\right)+\left(\frac{\cos\theta}{\sin\theta+1}+\frac{\cos^2\theta}{(\cos\theta+\sin\theta)^2}\right)\ d\theta=\int_0^{\pi/2}\frac{-(cos\theta-1)}{sin\theta}+\frac{1}{sin2\theta}+\frac{sin\theta-1}{cos\theta}=\int^{\pi/2}_{0}-\cot\theta+\csc\theta+\csc 2\theta+\tan\theta-\sec\theta d\theta$. Putting this into symbolab I get that it diverges. I don't know what I'm doing wrong.
 A: As it is stated the result is 0. To obtain $1$ it should be
$$\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}+\frac{x}{(x+y)^2}\right)dy.$$
Then
$$
\begin{align}\int_0^{\pi/2}&\left(\left(\frac{1}{\cos\theta+1}-\frac{\sin\theta}{(\cos\theta+\sin\theta)^2}\right)(-\sin\theta)+\left(\frac{1}{\sin\theta+1}+\frac{\cos\theta}{(\cos\theta+\sin\theta)^2}\right)\cos\theta \right)d\theta\\
&=\int_0^{\pi/2}\left(\frac{-\sin\theta}{\cos\theta+1}+\frac{\cos(\theta)}{\sin\theta+1}+\frac{1}{(\cos\theta+\sin\theta)^2}\right)d\theta\\
&=\left[\ln(\cos\theta+1)+\ln(\sin\theta+1)+\frac{\sin\theta}{\cos\theta+\sin\theta}\right]_0^{\pi/2}\\
&=\ln(2)+1-\ln(2)=1\end{align}$$
Note that in your work $(\cos\theta+\sin\theta)^2$ should be equal to $1+\sin(2\theta)$ and not $\sin(2\theta)$.
Actually, there is no need to compute the line integral because the given differential form is exact with $f(x,y)=\ln(x+1)+\ln(y+1)+\frac{y}{x+y}$ and
$$\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}+\frac{x}{(x+y)^2}\right)dy=f(0,1)-f(1,0)=1.$$
