$$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=I_t=?$$
Intento 1:
Si aplicamos la sustitución $x=1/u$, entonces nuestra integral será la siguiente;
$$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=-\displaystyle\int_{t}^{1/t}\dfrac1{u^2}\dfrac{du}{\left(\dfrac{1+u^2}{u^2}\right)\left(\dfrac{1+u^t}{u^t}\right)}=\displaystyle\int_{1/t}^t\dfrac{du}{\frac{(u^2+1)(u^t+1)}{u^t}}$$
Pero esto no tiene sentido:
$I_t=\displaystyle\int_{1/t}^t\dfrac{u^tdx}{(u^2+1)(u^t+1)}=\displaystyle\int_{1/t}^t\dfrac{x^tdx}{(x^2+1)(x^t+1)}=^?\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}$
Intento 2:
$$\dfrac{1}{(x^2+1)(x^t+1)}=\dfrac{Ax+B}{(x^2+1)}+\dfrac{h_{t-1}x^{t-1}+h_{t-2}x^{t-2}+...+h_1x+h_0}{(x^t+1)}$$$$\to$$ $$A+h_{t-1}=0\\B+h_{t-2}=0\\B+h_0=0\\A+h_1=0$$ son el otro $h_i$ $0$; $$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=\displaystyle\int_{1/t}^t\left(\dfrac{Ax+B}{(x^2+1)}+\dfrac{h_{t-1}x^{t-1}+h_{t-2}x^{t-2}+...+h_1x+h_0}{(x^t+1)}a\right)$$$$\to$$$$ \displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=\displaystyle\int_{1/t}^t\left(\dfrac{Ax+B}{(x^2+1)}-\dfrac{Ax^{t-1}+Bx^{t-2}+Ax+B}{(x^t+1)}\right)$$
Después de esto, he intentado aplicar la sustitución de las formas trigonométricas etc., pero yo no.