Bueno, en primer lugar podemos escribir:
$$\mathscr{F}\left(\text{x},\text{y}\right):=\frac{1+\text{x}+\text{y}^2+\text{x}^3+\dots}{1+\text{y}+\text{x}^2+\text{y}^3+\dots}=\frac{\sum\limits_{\text{n}=0}^\infty\text{x}^{2\text{n}+1}+\sum\limits_{\text{n}=0}^\infty\text{y}^{2\text{n}}}{\sum\limits_{\text{n}=0}^\infty\text{x}^{2\text{n}}+\sum\limits_{\text{n}=0}^\infty\text{y}^{2\text{n}+1}}\tag1$$
Ahora, podemos usar:
- Cuando $\left|\text{x}\right|<1$ : $$\sum_{\text{n}=0}^\infty\text{x}^{2\text{n}+1}=\frac{\text{x}}{1-\text{x}^2}\tag2$$
- Cuando $\left|\text{x}\right|<1$ : $$\sum_{\text{n}=0}^\infty\text{x}^{2\text{n}}=\frac{1}{1-\text{x}^2}\tag3$$
Así que, tenemos:
$$\mathscr{F}\left(\text{x},\text{y}\right)=\frac{\frac{\text{x}}{1-\text{x}^2}+\frac{1}{1-\text{y}^2}}{\frac{1}{1-\text{x}^2}+\frac{\text{y}}{1-\text{y}^2}}=\frac{\text{x}\cdot\left(\text{x}+\text{y}^2-1\right)-1}{\text{y}\cdot\left(\text{x}^2+\text{y}-1\right)-1}\tag4$$
Cuando $\left|\text{x}\right|<1\space\wedge\space\left|\text{y}\right|<1$
Para el $\text{x}$ integral que obtenemos:
$$\mathscr{I}_{\space\text{x}}\left(\text{y}\right):=\int_0^1\mathscr{F}\left(\text{x},\text{y}\right)\space\text{d}\text{x}=\int_0^1\frac{\text{x}\cdot\left(\text{x}+\text{y}^2-1\right)-1}{\text{y}\cdot\left(\text{x}^2+\text{y}-1\right)-1}\space\text{d}\text{x}=$$ $$\left(\text{y}^2-1\right)\int_0^1\frac{\text{x}-1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}+\frac{1}{\text{y}}\int_0^11\space\text{d}\text{x}\tag5$$
Así, obtenemos para las integrales indefinidas
- $$\int1\space\text{d}\text{x}=\text{x}+\text{C}_1\tag6$$
- $$\int\frac{\text{x}-1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}=$$ $$\int\frac{\text{x}}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}-\int\frac{1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}\tag7$$
- Sustituir $\text{u}:=\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1$ : $$\int\frac{\text{x}}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}=\frac{1}{2\cdot\text{y}}\int\frac{1}{\text{u}}\space\text{d}\text{u}=\frac{\ln\left|\text{u}\right|}{2\cdot\text{y}}+\text{C}_2\tag8$$
- Sustituir $\text{p}:=\frac{\text{x}\cdot\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}$ : $$\int\frac{1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}=\frac{1}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\int\frac{1}{1+\text{p}^2}\space\text{d}\text{p}=$$ $$\frac{\arctan\left(\text{p}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}+\text{C}_3\tag9$$
Así, para el $\text{x}$ integral que obtenemos:
$$\mathscr{I}_{\space\text{x}}\left(\text{y}\right)=\left(\text{y}^2-1\right)\cdot\left[\frac{\ln\left|\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1\right|}{2\cdot\text{y}}-\frac{\arctan\left(\frac{\text{x}\cdot\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\right]_0^1+\frac{1}{\text{y}}=$$ $$\left(\text{y}^2-1\right)\cdot\left\{\frac{\ln\left|\frac{\text{y}^2-1}{\text{y}^2-\text{y}-1}\right|}{2\cdot\text{y}}-\frac{\arctan\left(\frac{\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\right\}+\frac{1}{\text{y}}\tag{10}$$
