An integral for the new year

The $1 + x^2$ lends itself to the easy substitution $x = \tan \theta.$ Upon doing so, this transforms the integral accordingly:

$\int_0^\infty\dfrac{dx}{(1+x^2)(1+x^{2016})} \longrightarrow \int_0^{\frac{\pi}{2}} \dfrac{d\theta}{1 + \tan^{2016}\theta}$

From here, we'll break the integral up into two intervals, like so:

$\int_0^{\frac{\pi}{4}} \dfrac{d\theta}{1 + \tan^{2016}\theta} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{d\theta}{1 + \tan^{2016}\theta}$

Then, we'll write the second integral in terms of cotangent, rather than tangent: $\int_0^{\frac{\pi}{4}} \dfrac{d\theta}{1 + \tan^{2016}\theta} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\cot^{2016} \theta \, d\theta}{1 + \cot^{2016}\theta}$

Lastly, we'll make one more substitution for the second integral, namely $\theta = \frac{\pi}{2} - \beta$ , thus transforming the integral into

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\cot^{2016} \theta \, d\theta}{1 + \cot^{2016}\theta} \longrightarrow \int_{\frac{\pi}{4}}^{0} \dfrac{- \cot^{2016}(\frac{\pi}{2} - \beta) \, d\beta}{1 + \cot^{2016}(\frac{\pi}{2} - \beta)} = \int_{0}^\frac{\pi}{4} \dfrac{ \tan^{2016}(\beta) \, d\beta}{1 + \tan^{2016}(\beta)}$

This result can replace the second integral to give us the equivalent expression:

$\int_0^{\frac{\pi}{4}} \dfrac{d\theta}{1 + \tan^{2016}\theta} + \int_{0}^\frac{\pi}{4} \dfrac{ \tan^{2016}(\theta) \, d\theta}{1 + \tan^{2016}(\theta)} = \int_{0}^{\frac{\pi}{4}} 1 d\theta = \frac{\pi}{4} \quad \square$

well I took a slightly different approach $I = \int_0^\infty \frac{dx}{(x^2 + 1)(x^{2016} + 1)} \Longrightarrow (1)$ set x = 1/y then $dx = \frac{-dy}{y^2}$ and the new bounds are $y = \lim_{x \rightarrow 0^+} \frac{1}{x}= \infty$ and $y = \lim_{x \rightarrow \infty} \frac{1}{x}= 0$

so the integral becomes $I = \int_\infty^0 \frac{\frac{-dy}{y^2}}{\frac{(y^2 + 1)(y^{2016} + 1)}{y^2 \times y^{2016}}} = \int_0^\infty \frac{y^{2016} dy}{(y^2 + 1)(y^{2016} + 1)} = \int_0^\infty \frac{x^{2016} dx}{(x^2 + 1)(x^{2016} + 1)} \Longrightarrow (2)$ now add (1) and (2) $\Longrightarrow 2 \times I = \int_0^\infty \frac{dx}{1 + x^2} = tan^{-1} \infty - tan^{-1} 0 = \frac{\pi}{2} \Longrightarrow I = \frac{\pi}{4}$

We can substitute $u=\tan\theta,$ motivated by how there would be a $\sec^2\theta\text{ }d\theta$ in the numerator and a $\sec^2\theta$ in the denominator. This will transform the integral into the following: $\int_0^\frac{\pi}{2}\frac{d\theta}{1+\tan^{2016}\theta}$ Let's examine the integral $I=\displaystyle\int_0^\frac{\pi}{2}\frac{d\theta}{1+\tan^c\theta}.$ We can make use of the following identity: $\int_a^bf(x)\text{ }dx=\int_a^bf(a+b-x)\text{ }dx$ We find the following. $\int_0^\frac{\pi}{2}\frac{d\theta}{1+\tan^c\theta}=\int_0^\frac{\pi}{2}\frac{d\theta}{1+\cot^c\theta}$ Summing these, we find this. $2I=\int_0^\frac{\pi}{2}\left(\frac{1}{1+\tan^c\theta}+\frac{1}{1+\cot^c\theta}\right)d\theta=\int_0^\frac{\pi}{2}\left(\frac{1}{1+\tan^c\theta}+\frac{\tan^c\theta}{\tan^c\theta+1}\right)d\theta=\int_0^\frac{\pi}{2}\frac{1+\tan^c\theta}{1+\tan^c\theta}d\theta$

This simplifies to $2I$ being equal to $\displaystyle\int_0^\frac{\pi}{2}1\text{ }d\theta.$ Obviously, $2I=\frac{\pi}{2},$ so $I=\frac{\pi}{4}$ and $A=\boxed{4}.$