calculate-dx-x-2-x-1-3-

Question Number 65129 by turbo msup by abdo last updated on 25/Jul/19

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{3}}

Commented by mathmax by abdo last updated on 26/Jul/19

l e t A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{3}} l e t t u r n t o m a g i c a l n u m b e r i

w e h a v e x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4} a n d c h a n g . x + \frac{1}{2} = \frac{\sqrt{3}}{2} t g i v e

A = \int_{- \infty}^{+ \infty} \frac{1}{{(\frac{3}{4})}^{3} {(t^{2} + 1)}^{3}} \frac{\sqrt{3}}{2} d t = \frac{4^{3}}{3^{3}} \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{3}}

= \frac{32 \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}}

t h e p o l e s o f φ a r e \bar{+} i (t r i p l e s) r e s i d u s t h e o r e m g i v e

\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)}

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

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

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{3}{16 i} = \frac{3 π}{8} \Rightarrow A = \frac{32 \sqrt{3}}{27} \frac{3 π}{8} = \frac{3 π \sqrt{3} . 8.4}{3.9 . 8}

A = \frac{4 π \sqrt{3}}{9} .

Answered by MJS last updated on 29/Jul/19

\int \frac{d x}{{(x^{2} + x + 1)}^{3}} =

{Ostrogradski}^{'} s Method

\int \frac{p (x)}{q (x)} = \frac{p_{1} (x)}{q_{1} (x)} + \int \frac{p_{2} (x)}{q_{2} (x)}

p (x) = 1 q (x) = {(x^{2} + x + 1)}^{3}

q_{1} (x) = {(x^{2} + x + 1)}^{2} q_{2} (x) = x^{2} + x + 1

p_{1} (x) = \frac{(2 x + 1) (2 x^{2} + 2 x + 3)}{6} p_{2} = \frac{2}{3}

= \frac{(2 x + 1) (2 x^{2} + 2 x + 3)}{6 {(x^{2} + x + 1)}^{2}} + \frac{2}{3} \int \frac{d x}{x^{2} + x + 1} =

= \frac{(2 x + 1) (2 x^{2} + 2 x + 3)}{6 {(x^{2} + x + 1)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1)) + C

\underset{- \infty}{\int^{+ \infty}} \frac{d x}{{(x^{2} + x + 1)}^{3}} = \frac{4 \sqrt{3}}{9} π

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