Closed contour integral: $\int_{\mathbb{c}}\frac{ z}{2z^{2}+1} dz$ where the contour is the unit circle first and foremost please excuse my English.
given $∫_c \frac{{z}}{2z^{2}+1}dz$  where the contour is the unit circle. so c = $e^{it}$ from 0 to $2\pi$. 
since the contour is the unit circle we can say that $f(z(t)) = \frac{e^{it}}{2*(e^{it})^{2}+1}$ and $z'(t) = i*e^{it}$
We know that $\int_a^{b} f(z(t))*z'(t) = \int _c f(z) dz$
so then we just substitute what we know and we get:$$ \int _0 ^{2\pi } \frac{e^{it}}{2*(e^{it})^{2}+1} *  i*e^{it} dt $$
$$ i\int _0 ^{2\pi } \frac{e^{2it}}{2e^{2it}+1} dt $$
we let $ u = 2e^{2it}+1 $ and $du = 4ie^{2it}$ and we get : 
$$i\int _0 ^{2\pi } \frac{e^{2it}}{u} \frac{du}{4ie^{2it}}$$ 
$$ \frac{1}{4}\int _0 ^{2\pi } \frac{1}{u}du $$
we solve and see that:
$$ i\int _0 ^{2\pi } \frac{e^{2it}}{2e^{2it}+1} dt = \frac{1}{4}(log (1+2e^{4\pi i})-log(1+2e^{0 i})) = 0 $$
is this correct? this was a problem on my final and when I computed this contour integral on wolfram alpha I got $\pi$i?
any explanation would by much appreciate it. I understand I could have done this problem with Cauchy's Integral Formula. Our class did not get up to residue calculus since this is an undergraduate course. Many thanks in advance. 

No, the answer is $\pi i$. Since 
$$\frac{z}{2z^2 + 1} = \frac{1/4}{z - i/\sqrt{2}} + \frac{1/4}{z + i/\sqrt{2}}$$
then 
$$\int_c \frac{z}{2z^2 + 1}\, dz = \frac{1}{4}\int_c \frac{1}{z - i\sqrt{2}}\, dz + \frac{1}{4}\int_c \frac{1}{z + i/\sqrt{2}}\, dz = \frac{1}{4}(2\pi i) + \frac{1}{4}(2\pi i) = \pi i$$
using Cauchy's integral formula in the second to last step.
