TITLE: Calculate $\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}$
QUESTION [6 upvotes]: $$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=I_t=?$$

Attempt 1:
If we apply substitution $x=1/u$, then our integral will be like following;
$$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=-\displaystyle\int_{t}^{1/t}\dfrac1{u^2}\dfrac{du}{\left(\dfrac{1+u^2}{u^2}\right)\left(\dfrac{1+u^t}{u^t}\right)}=\displaystyle\int_{1/t}^t\dfrac{du}{\frac{(u^2+1)(u^t+1)}{u^t}}$$
But this doesn't make any sense:
$I_t=\displaystyle\int_{1/t}^t\dfrac{u^tdx}{(u^2+1)(u^t+1)}=\displaystyle\int_{1/t}^t\dfrac{x^tdx}{(x^2+1)(x^t+1)}=^?\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}$
Attempt 2:
$$\dfrac{1}{(x^2+1)(x^t+1)}=\dfrac{Ax+B}{(x^2+1)}+\dfrac{h_{t-1}x^{t-1}+h_{t-2}x^{t-2}+...+h_1x+h_0}{(x^t+1)}$$$$\to$$ $$A+h_{t-1}=0\\B+h_{t-2}=0\\B+h_0=0\\A+h_1=0$$
other $h_i$ are $0$;
$$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=\displaystyle\int_{1/t}^t\left(\dfrac{Ax+B}{(x^2+1)}+\dfrac{h_{t-1}x^{t-1}+h_{t-2}x^{t-2}+...+h_1x+h_0}{(x^t+1)}a\right)$$$$\to$$$$ \displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=\displaystyle\int_{1/t}^t\left(\dfrac{Ax+B}{(x^2+1)}-\dfrac{Ax^{t-1}+Bx^{t-2}+Ax+B}{(x^t+1)}\right)$$
After this, I tried to apply substitution of trigonometric forms etc., but I failed.

REPLY [12 votes]: Hint: You've shown $$I = \int_{1/t}^t \frac{x^t \, \mathrm{d}x}{(x^2+1)(x^t+1)} = \int_{1/t}^t \frac{\mathrm{d}x}{(x^2+1)(x^t+1)}$$
Now add them together to get $$2I = \int_{1/t}^t \frac{x^t + 1}{(x^2+1)(x^t+1)} \, \mathrm{d}x = \int_{1/t}^t \frac{\mathrm{d}x}{1+x^2}$$

As per Dr. MV's excellent comment, this technique works for any integral of the form $\int_{1/a}^a \frac{\mathrm{d}x}{(x^2+1)(x^b+1)}$, that is having both the upper limit be $t$ and the exponent in the integrand be $t$ as well has no relevance. 
In fact, your (clever) substitution is what shows us that the value of the exponent in $(x^t+1)$ is entirely irrelevant to the problem.