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. 

The solution presented in the OP might appear paradoxical.  The substitution $u=1+e^{i2t}$ seems both natural and innocuous.  Yet, inappropriate application of the substitution led to the integral
$$\int_{t=0}^{t=2\pi}\frac{1}{u}\,du=0$$
So, what went wrong here?


In THIS ANSWER, I showed that for any rectifiable curve $\gamma$ on $\mathbb{C}\setminus\{0\}$, that begins at $1$ and ends at $re^{i\theta}$
$$\int_\gamma \frac1z\,dz=\log(r)+i(\theta +2\pi k) \tag 1$$
where $k$ is the net number of times $\gamma$ crosses the positive real axis from the fourth quadrant to the first quadrant.  (For additional explanation, SEE THIS ANSWER).
Note that $(1)$ can be easily generalized by replacing the starting point at $z=x$.  In that case, we simply replace $\log(r)$ with $\log(r/x)$.


To make things clear, we apply $(1)$ to the problem of interest.  Observe that for $t\in [0,2\pi]$, the substitution $u=1+e^{2it}$ results in a contour $\gamma$ defined by $|u-1|=2$, that has one net crossing (at $t=\pi$) of the positive real axis in the $u$-plane.  
Then, using the aforementioned modification of $(1)$, we find that with $x=r=3$, $\theta=2\pi$, and $k=1$
$$\int_{|u-1|=2}\frac1u\,du =i4\pi$$
Finally, we have
$$\begin{align}
\oint_{|z|=1} \frac{z}{2z^2+1}\,dz&=i\int_0^{2\pi}\frac{e^{i2t}}{2e^{i2t}+1}\,dt\\\\
&=\frac14 \oint_{|u-1|=2}\frac1u\,du\\\\
&=i\pi
\end{align}$$
as was to be shown!

