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$$\displaystyle\int_{1/t}^t\dfrac{dx}{(x^2+1)(x^t+1)}=I_t=?$$

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.

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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.

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    $\begingroup$ Just beat me :) but yes, this is definitely the way to go. Apparently the presence of the $x^t$ in the numerator scared the OP? $\endgroup$ – Brevan Ellefsen May 3 '17 at 21:59
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    $\begingroup$ Indeed, I can understand why. @OP It may seem counter intuitive that $\int_a^b \frac{f}{g} = \int_a^b \frac{1}{g}$, but remember that two different functions can integrate (over a region) to the same thing, your result isn't wrong in any sense. $\endgroup$ – Zain Patel May 3 '17 at 22:02
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    $\begingroup$ (+1) You might consider adding a note that exposes that the having $t$ be both the exponent and the upper integration limit is a red herring. The approach works for any exponent on $x$ provided the limits are reciprocals of each other (i.e., $\int_a^{1/a} \frac{1}{(x^2+1)(x^b+1)}\,dx$). $\endgroup$ – Mark Viola May 3 '17 at 22:57
  • $\begingroup$ @Dr.MV Done, thanks very much for the suggestion! $\endgroup$ – Zain Patel May 3 '17 at 23:01
  • $\begingroup$ Zain, appreciative of the reference. ;-) $\endgroup$ – Mark Viola May 4 '17 at 2:35

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