Question #6f0de

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Question #6f0de

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Answer 1 · 2017-11-22T19:02:34

$\int_{- \infty}^{\infty} \frac{x^{2}}{x^{6} + 9} \cdot d x = \frac{π}{9}$ $int_-oo^oo x^2/(x^6+9)*dx=pi/9$

Explanation:

$\int_{- \infty}^{\infty} \frac{x^{2}}{x^{6} + 9} \cdot d x$ $int_-oo^oo x^2/(x^6+9)*dx$

= $\frac{1}{3} \cdot \int_{- \infty}^{\infty} \frac{3 x^{2} \cdot d x}{{(x^{3})}^{2} + 9}$ $1/3*int_-oo^oo (3x^2*dx)/[(x^3)^2+9]$

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

$\frac{1}{3} \cdot \int_{- \infty}^{\infty} \frac{d y}{y^{2} + 9}$ $1/3*int_-oo^oo dy/(y^2+9)$

= $\frac{2}{3} \cdot \int_{0}^{\infty} \frac{d y}{y^{2} + 9}$ $2/3*int_0^oo dy/(y^2+9)$

= $\frac{2}{9} \cdot {[arctan (\frac{y}{3})]}_{0}^{\infty}$ $2/9*[arctan(y/3)]_0^oo$

= $\frac{2}{9} \cdot [arctan (\infty) - arctan (0)]$ $2/9*[arctan(oo)-arctan(0)]$

= $\frac{2}{9} \cdot (\frac{π}{2} - 0)$ $2/9*(pi/2-0)$

= $\frac{π}{9}$ $pi/9$

Answer 2 · 2017-11-22T23:56:50

2nd way: I didn't use symmetry

$\int_{- \infty}^{\infty} \frac{x^{2}}{x^{6} + 9} \cdot d x$ $int_-oo^oo x^2/(x^6+9)*dx$

= $\frac{1}{3} \cdot \int_{- \infty}^{\infty} \frac{3 x^{2} \cdot d x}{{(x^{3})}^{2} + 9}$ $1/3*int_-oo^oo (3x^2*dx)/[(x^3)^2+9]$

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

$\frac{1}{3} \cdot \int_{- \infty}^{\infty} \frac{d y}{y^{2} + 9}$ $1/3*int_-oo^oo dy/(y^2+9)$

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

$\frac{1}{3} \cdot \int_{- \infty}^{a} \frac{d y}{y^{2} + 9} + \frac{1}{3} \cdot \int_{a}^{\infty} \frac{d y}{y^{2} + 9}$ $1/3*int_-oo^a dy/(y^2+9)+1/3*int_a^oo dy/(y^2+9)$

= $\frac{1}{9} \cdot {[arctan (\frac{y}{3})]}_{- \infty}^{a} + \frac{1}{9} \cdot {[arctan (\frac{y}{3})]}_{a}^{\infty}$ $1/9*[arctan(y/3)]_-oo^a+1/9*[arctan(y/3)]_a^oo$

= $\frac{1}{9} \cdot [arctan (\frac{a}{3}) - arctan (- \infty)] + \frac{1}{9} \cdot [arctan (\infty) - arctan (\frac{a}{3})]$ $1/9*[arctan(a/3)-arctan(-oo)]+1/9*[arctan(oo)-arctan(a/3)]$

= $\frac{1}{9} \cdot [arctan (\infty) - arctan (- \infty)]$ $1/9*[arctan(oo)-arctan(-oo)]$

= $\frac{1}{9} \cdot [\frac{π}{2} - (- \frac{π}{2})]$ $1/9*[pi/2-(-pi/2)]$

= $\frac{π}{9}$ $pi/9$

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Question #6f0de

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