Question #6f0de

Nov 22, 2017

${\int}_{-} {\infty}^{\infty} {x}^{2} / \left({x}^{6} + 9\right) \cdot \mathrm{dx} = \frac{\pi}{9}$

Explanation:

${\int}_{-} {\infty}^{\infty} {x}^{2} / \left({x}^{6} + 9\right) \cdot \mathrm{dx}$

=$\frac{1}{3} \cdot {\int}_{-} {\infty}^{\infty} \frac{3 {x}^{2} \cdot \mathrm{dx}}{{\left({x}^{3}\right)}^{2} + 9}$

After using $y = {x}^{3}$ an $\mathrm{dy} = 3 {x}^{2} \cdot \mathrm{dx}$ transformation, this integral became

$\frac{1}{3} \cdot {\int}_{-} {\infty}^{\infty} \frac{\mathrm{dy}}{{y}^{2} + 9}$

=$\frac{2}{3} \cdot {\int}_{0}^{\infty} \frac{\mathrm{dy}}{{y}^{2} + 9}$

=$\frac{2}{9} \cdot {\left[\arctan \left(\frac{y}{3}\right)\right]}_{0}^{\infty}$

=$\frac{2}{9} \cdot \left[\arctan \left(\infty\right) - \arctan \left(0\right)\right]$

=$\frac{2}{9} \cdot \left(\frac{\pi}{2} - 0\right)$

=$\frac{\pi}{9}$

Nov 22, 2017

2nd way: I didn't use symmetry

${\int}_{-} {\infty}^{\infty} {x}^{2} / \left({x}^{6} + 9\right) \cdot \mathrm{dx}$

=$\frac{1}{3} \cdot {\int}_{-} {\infty}^{\infty} \frac{3 {x}^{2} \cdot \mathrm{dx}}{{\left({x}^{3}\right)}^{2} + 9}$

After using $y = {x}^{3}$ an $\mathrm{dy} = 3 {x}^{2} \cdot \mathrm{dx}$ transformation, this integral became

$\frac{1}{3} \cdot {\int}_{-} {\infty}^{\infty} \frac{\mathrm{dy}}{{y}^{2} + 9}$

Now I divided range of integral. First of it from $- \infty$ to $a$ and second of it from $a$ to $\infty$.

$\frac{1}{3} \cdot {\int}_{-} {\infty}^{a} \frac{\mathrm{dy}}{{y}^{2} + 9} + \frac{1}{3} \cdot {\int}_{a}^{\infty} \frac{\mathrm{dy}}{{y}^{2} + 9}$

=$\frac{1}{9} \cdot {\left[\arctan \left(\frac{y}{3}\right)\right]}_{-} {\infty}^{a} + \frac{1}{9} \cdot {\left[\arctan \left(\frac{y}{3}\right)\right]}_{a}^{\infty}$

=$\frac{1}{9} \cdot \left[\arctan \left(\frac{a}{3}\right) - \arctan \left(- \infty\right)\right] + \frac{1}{9} \cdot \left[\arctan \left(\infty\right) - \arctan \left(\frac{a}{3}\right)\right]$

=$\frac{1}{9} \cdot \left[\arctan \left(\infty\right) - \arctan \left(- \infty\right)\right]$

=$\frac{1}{9} \cdot \left[\frac{\pi}{2} - \left(- \frac{\pi}{2}\right)\right]$

=$\frac{\pi}{9}$