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Evaluate

$$ \int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx \, \, \, (\phi>0) $$

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Solution

$$ \small \int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx = \int_0^{1} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx + \int_1^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx $$

Notice that with the substitution $y=\frac{1}{x}$, $$\begin{aligned} \int_1^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx &= \int_1^0 \frac{1}{\left( 1+\frac{1}{y^2}\right)\left(1+\frac{1}{y^{\phi}} \right)}\cdot \left(\frac{-1}{y^2} \right)\, dy\\ &=\int_0^1\frac{y^{\phi}}{(1+y^2)(1+y^{\phi})}\, dy \end{aligned}$$

Therefore

$$\begin{aligned} \int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx &= \int_0^1\frac{1+x^{\phi}}{(1+x^2)(1+x^{\phi})}\, dx \\ &= \int_0^1 \frac{1}{1+x^2}\, dx \\ &= \arctan x|_0^1 \\ &= \frac{\pi}{4} \end{aligned}$$

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Another way would be to make the substitution $x\mapsto\tan x$ so that$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\frac {\mathrm dx}{1+\tan^nx}\\ & =\int\limits_0^{\pi/2}\mathrm dx-\int\limits_0^{\pi/2}\mathrm dx\,\frac {\tan^n x}{1+\tan^nx}\\ & =\frac {\pi}2-\int\limits_0^{\pi/2}\mathrm dx\,\frac 1{1+\cot^nx}\end{align*}$$Now use the identity that$$\cot x=\tan\left(\frac {\pi}2-x\right)$$to get that$$\begin{align*}\mathfrak{I} & =\frac {\pi}2-\int\limits_0^{\pi/2}\frac {\mathrm dx}{1+\tan^nx}\\ & =\frac {\pi}2-\mathfrak{I}\end{align*}$$

Now isolate $\mathfrak{I}$ to get that$$\int\limits_0^{\infty}\frac {\mathrm dx}{(1+x^2)(1+x^n)}\color{blue}{=\frac {\pi}4}$$

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