find-dx-x-2-2x-4-3-

Question Number 93873 by mathmax by abdo last updated on 15/May/20

f i n d \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} + 2 x + 4)}^{3}}

Commented by mathmax by abdo last updated on 16/May/20

I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 x + 4)}^{3}} \Rightarrow I = \int_{- \infty}^{+ \infty} \frac{d x}{{({(x + 1)}^{2} + 3)}^{3}}

=_{x + 1 = \sqrt{3} t} \int_{- \infty}^{+ \infty} \frac{\sqrt{3} d t}{27 {(t^{2} + 1)}^{3}} = \frac{\sqrt{3}}{27} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{3}}

l e t φ (z) = \frac{1}{{(z^{2} + 1)}^{3}} \Rightarrow φ (z) = \frac{1}{{(z - i)}^{3} {(z + i)}^{3}} s o t h e p o l e s o f φ a r e

i a n d - i (t r i p l e s) a n d \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}

= \frac{1}{2} {l i m}_{z \to i} {{(z + i)}^{- 3}}^{(2)} = \frac{1}{2} {l i m}_{z \to i} {- 3 {(z + i)}^{- 4}}^{(1)}

= \frac{1}{2} {l i m}_{z \to i} 12 {(z + i)}^{- 5} = 6 {(2 i)}^{- 5} = \frac{6}{{(2 i)}^{5}} = \frac{6}{32 i} = \frac{3}{16 i} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{3}{16 i} = \frac{3 π}{8} \Rightarrow I = \frac{\sqrt{3}}{27} \times \frac{3 π}{8} = \frac{π \sqrt{3}}{72}

★ I = \frac{π \sqrt{3}}{72} ★

Answered by Ar Brandon last updated on 15/May/20

L_{n} = \int \frac{Ax + B}{{(x^{2} + px + q)}^{n}} dx \Rightarrow L_{n} = (B - \frac{Ap}{2}) \int_{- \infty}^{+ \infty} \frac{1}{{[{(x + \frac{p}{2})}^{2} + {(\sqrt{q - \frac{p^{2}}{4}})}^{2}]}^{n}} dx

A = 0, B = 1, p = 2, q = 4, n = 3

\Rightarrow L_{3} = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + 2 x + 4)}^{3}} = \int_{- \infty}^{+ \infty} \frac{dx}{{[{(x + 1)}^{2} + {(\sqrt{3})}^{2}]}^{3}}

J_{n} = \int_{- \infty}^{+ \infty} \frac{1}{{[x^{2} + u^{2}]}^{n}} dx \Rightarrow J_{n} = \frac{2 n - 3}{{2 u}^{2} (n - 1)} J_{n - 1}

J_{1} = \int_{- \infty}^{+ \infty} \frac{dx}{{(x + 1)}^{2} + {(\sqrt{3})}^{2}} = \frac{\sqrt{3}}{3} ⌊ \tan^{- 1} [\frac{x + 1}{\sqrt{3}}] ⌋_{- \infty}^{+ \infty} = \frac{π \sqrt{3}}{3}

J_{2} = \frac{π \sqrt{3}}{18} \Rightarrow J_{3} = \frac{3}{12} \cdot \frac{π \sqrt{3}}{18} = \frac{π \sqrt{3}}{72}

\int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + 2 x + 4)}^{3}} = \frac{π \sqrt{3}}{72}

Commented by Ar Brandon last updated on 15/May/20

In case of necessity please check my solution to Q.93639 for Ln and Jn ��

Commented by mathmax by abdo last updated on 16/May/20

t h a n k y o u s i r .

Commented by Ar Brandon last updated on 16/May/20

cheers ��