Computation integral with complex analysis 
Compute $$\frac{1}{2 \pi i} \int_{\partial B_1(0)} \frac{6z^{98}}{23z^{99}-2z^{81}+z^4-7}dz$$

Im clueless. I would use the residue theorem, but factorising the denominator is not an option, I proved with Rouche that all 99 singularities are inside $B_1(0)$. None of them is removable. The numerator looks suspicious, but I don't really know what to do with it: one idea would be to use the argument principle but the integrand is not in the wished form.
 A: Since every singularity lies in $B_1(0)$, then$$\oint_{|z|=R}\frac{6z^{98}}{23z^{99}-2z^{81}+z^4-7}\,\mathrm dz$$is independent of the choice of $R$, as long as $R\geqslant1$. And\begin{multline}\lim_{R\to\infty}\oint_{|z|=R}\frac{6z^{98}}{23z^{99}-2z^{81}+z^4-7}-\frac6{23z}\,\mathrm dz=\\=\lim_{R\to\infty}\oint_{|z|=R}\frac{6 \left(2 z^{81}-z^4+7\right)}{23\left(23 z^{100}-2z^{82}+z^5-7z\right)}\,\mathrm dz=0\end{multline}So,$$\frac1{2\pi i}\oint_{|z|=R}\frac{6z^{98}}{23z^{99}-2z^{81}+z^4-7}\,\mathrm dz=\frac6{23}.$$
A: The fact that the numerator and denominator's powers are off by $1$ ($98$ vs $99$) is very suggestive of using the change of variables $z=\frac{1}{\zeta}$. This is further reinforced by the fact that all the poles lie inside $|z|=1$. If you make this change of variables then (be careful with orientation of the curve)
\begin{align}
\frac{1}{2\pi i}\int_{|z|=1}f(z)\,dz&=-\frac{1}{2\pi i}\int_{|\zeta|=1}f\left(\frac{1}{\zeta}\right)\cdot \left(-\frac{d\zeta}{\zeta^2}\right)\\
&=-\frac{1}{2\pi i}\int_{|\zeta|=1}\frac{6\zeta}{23-2\zeta^{18}+\zeta^{95}-7\zeta^{99}}\cdot\frac{-d\zeta}{\zeta^2}\\
&=\frac{1}{2\pi i}\int_{|\zeta|=1}\frac{6}{23-2\zeta^{18}+\zeta^{95}-7\zeta^{99}}\cdot\frac{1}{\zeta}\,d\zeta
\end{align}
Now by Rouche's theorem, you can once again verify (compare with the constant function $23$) that the denominator has no zeros inside $|\zeta|<1$. Hence, the pole is only at $\zeta=0$ with a residue of $\frac{6}{23}$.
