I used partial fraction decomposition and wrote $$\int_C\frac{z}{z^2+1}\,dz=\frac12\int_c\frac{1}{z+i}\,dz+\frac12\int_C\frac{1}{z-i}\,dz$$ Here is an image for $C$.
Using the Extended Cauchy-Goursat Theorem, I wrote
$$ \begin{align} \int_C\frac{z}{z^2+1}\,dz&=\frac12\int_{C_1}\frac{dz}{z+i}+\frac12\int_{C_2}\frac{dz}{z+i}+\frac12\int_{C_1}\frac{dz}{z-i}+\frac12\int_{C_2}\frac{dz}{z-i}\\[6pt] \int_C\frac{z}{z^2+1}\,dz&=\frac12(0)+\frac12(2\pi\,i)+\frac12(2\pi\,i)+\frac12(0)\\[6pt] \int_C\frac{z}{z^2+1}\,dz&=2\pi\,i \end{align} $$
Next, I tried to use a parametrization of the circle, $$z(t)=2e^{it}\qquad\Rightarrow\qquad z'(t)\,dt=2ie^{it}\,dt\qquad\text{where}\qquad-\pi\le t\le \pi$$
Substituting these, I get: $$ \begin{array}{tcl} \int_C\dfrac{z}{z^2+1}\,dz&=&\int_{-\pi}^\pi\dfrac{2e^{it}}{(2e^{it})^2+1}(2ie^{it}\,dt)\\ \int_C\dfrac{z}{z^2+1}\,dz&=&\int_{-\pi}^\pi\dfrac{4ie^{i2t}}{4e^{i2t}+1}\,dt\\ \int_C\dfrac{z}{z^2+1}\,dz&=&\dfrac12\int_{-\pi}^\pi\dfrac{8ie^{i2t}\,dt}{4e^{i2t}+1}\,dt \end{array} $$
Now, I used $$u=4e^{i2t}+1\qquad\Rightarrow\qquad du=8ie^{i2t}\,dt$$ Note that $$t=-\pi\qquad\Rightarrow\qquad u=4e^{i2(-\pi)}+1=4(1)+1=5$$ Similarly, $t=\pi\rightarrow u=5$. Now, the integral becomes $$\frac12\int_5^5\frac{du}{u}$$ Now, I think this might equal zero, which is a different answer. What have I done wrong or what law am I disobeying?
Here is what Matlab returns.
syms z
clear f
f(z)=z/(z^2+1)
syms t real
z(t)=2*exp(i*t);
int(f(z(t))*diff(z(t),t),t,-pi,pi)
Ans:
$2\pi i$
Update Thanks to Pythagoras, I think I understand what is happening. Here is some Matlab code that demonstrates his statement that arg increases by $4\pi$.
t=linspace(0,2*pi);
z=4*exp(i*2*t)+1; u=real(z); v=imag(z);
plot(z,'Color','b')
axis equal,axis([-4,6,-5,5])
grid on
ax=gca; ax.XAxisLocation='origin'; ax.YAxisLocation='origin';
hold on
n=length(t);
for k=1:n
h=quiver(0,0,u(k),v(k),0,'Color','r','LineWidth',1);
pause(0.1);
delete(h)
end
quiver(0,0,5,0,0,'Color','r','LineWidth',1);
First of all, you meant $t=-\pi$ and $t=\pi$, not $t=-2\pi$, etc. When $t$ goes through a period of $2\pi$, as you can check $u$ wraps around $0$ twice, so $\arg u$ changes by $4\pi$. Note that $\int \frac 1 u~du$ is not single valued; in general it is given by $\ln |u|+i\arg u$, so by the above remark $$\frac 12\int \frac 1u~du=\frac 12\cdot 4\pi i=2\pi i.$$