Complex integral involving logarithm

Question

I've been working on this integral for quite a while and I think I've been able to progress but now I'm stuck. So I have to prove that

$$\int_C f(z)\ dz =\int_C\frac{2z}{(1+z^2)\log(2+z^2)}dz =\pi i.$$

Where $C$ is a contour from $[1,p]$ to $[p,-1]$ and $p=\frac{6i}5$.

---

My attempt :

1. So I have noticed that $f(-z)=-f(z)$ and so $f$ is odd and the contour can be completed (triangle) to be closed since $\int_Df(z)dz=0$ where $D$ is a line segment from $-1$ to $1$.
2. Clearly the only singularity in the new contour (the triangle) is $i$.

So this is where I get stuck. I've tried to use Cauchy's integral formula but the $\log(2+z^2)$ gives me hard time.

I would appreciate a hint. (Residue theorem is not allowed, almost everything else is).

Answer 1

Suggestion: Substitute $w = 1+z^2$ to simplify the situation. The image of $C$ under that substitution becomes a closed curve $\gamma$ in the half-plane $H = \{ w : \operatorname{Re} w > -1\}$ winding once around $0$, and the integral

$$\int_\gamma \frac{dw}{w\log (1+w)}.$$

Now we want to transform it into a way so that we can use the Cauchy integral formula (which is a special case of the residue theorem, but one that seems to be allowed). For that, consider the function

$$f(w) = \begin{cases}\qquad 1 &, w = 0\\ \dfrac{w}{\log (1+w)} &, w\neq 0 \end{cases}$$

which is holomorphic on $H$. By the integral formula,

$$\int_\gamma \frac{dw}{w\log (1+w)} = \int_\gamma \frac{f(w)}{w^2}\,dw = 2\pi i f'(0).$$

Of course you can play the same game without the substitution after closing $C$ with

$$F(z) = \begin{cases} \dfrac{2z(z-i)}{(z+i)\log (2+z^2)} &, z \neq i\\ \qquad \dfrac{1}{2i} &, z = i,\end{cases}$$

but it's easier to compute with the substitution.