calculate-1-dx-x-2-1-2-x-2-4-2-

Question Number 105230 by mathmax by abdo last updated on 27/Jul/20

calculate \int_{1}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}}

Answered by 1549442205PVT last updated on 27/Jul/20

F = \int {[\frac{1}{4} (\frac{1}{x^{2}} - \frac{1}{x^{2} + 4})]}^{2} dx

16 F = \int [\frac{1}{x^{4}} + \frac{1}{{(x^{2} + 4)}^{2}} - \frac{2}{x^{2} (x^{2} + 4)}] dx

= \int \frac{dx}{x^{4}} + \int (\frac{1}{2 (x^{2} + 4)} - \frac{1}{{2 x}^{2}}) dx + \int \frac{dx}{{(x^{2} + 4)}^{2}}

= - \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{1}{2} \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + I_{2}

= - \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{1}{4} \tan^{- 1} (\frac{x}{2}) + \frac{1}{2.4} . \frac{x}{(x^{2} + 4)} + \frac{1}{4} . \frac{1}{2} . \frac{1}{2} \tan^{- 1} (\frac{x}{2})

= - \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{x}{8 (x^{2} + 4)} + \frac{5}{16} \tan^{- 1} (\frac{x}{2})

F = \frac{1}{16} {- \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{x}{8 (x^{2} + 4)} + \frac{5}{16} \tan^{- 1} (\frac{x}{2})}

I = \int_{1}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}} = F (+ \infty) - F (1)

= \frac{1}{16} {\frac{5}{16} \times \frac{π}{2} - [\frac{23}{120} + \frac{5}{16} \tan^{- 1} (\frac{1}{2})]}

Commented by abdomsup last updated on 28/Jul/20

t h a n k y o u s i r

Commented by 1549442205PVT last updated on 28/Jul/20

Thank, you are welcome .

Answered by mathmax by abdo last updated on 29/Jul/20

I = \int_{1}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}}

we have F (x) = \frac{1}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}} = \frac{1}{9} {(\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4})}^{2}

= \frac{1}{9} {\frac{1}{{(x^{2} + 1)}^{2}} - \frac{2}{(x^{2} + 1) (x^{2} + 4)} + \frac{1}{{(x^{2} + 4)}^{2}}} \Rightarrow

9 F (x) = \frac{1}{{(x^{2} + 1)}^{2}} - \frac{2}{3} (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4}) + \frac{1}{{(x^{2} + 4)}^{2}}

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

9 \int_{1}^{\infty} F (x) = \int_{1}^{\infty} \frac{dx}{{(1 + x^{2})}^{2}} - \frac{2}{3} \int_{1}^{\infty} \frac{dx}{1 + x^{2}} + \frac{2}{3} \int_{1}^{\infty} \frac{dx}{x^{2} + 4} + \int_{1}^{\infty} \frac{dx}{{(x^{2} + 4)}^{2}}

we have \int_{1}^{\infty} \frac{dx}{{(1 + x^{2})}^{2}} =_{x = tant} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{(1 + \tan^{2} t)}{{(1 + \tan^{2} t)}^{2}} dt

= \int_{\frac{π}{4}}^{\frac{π}{2}} \cos^{2} t dt = \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} (1 + \cos (2 t)) dt = \frac{π}{8} + \frac{1}{4} {[\sin (2 t)]}_{\frac{π}{4}}^{\frac{π}{2}}

= \frac{π}{8} - \frac{1}{4} and \int_{1}^{\infty} \frac{dx}{1 + x^{2}} = {[arctanx]}_{1}^{\infty} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}

\int_{1}^{\infty} \frac{dx}{x^{2} + 4} =_{x = 2 t} \int_{\frac{1}{2}}^{\infty} \frac{2 dt}{4 (1 + t^{2})} = \frac{1}{2} (\frac{π}{2} - \arctan (\frac{1}{2}))

= \frac{π}{4} - \frac{1}{2} (\frac{π}{2} - \arctan 2) = \frac{\arctan 2}{2}

\int_{1}^{\infty} \frac{dx}{{(x^{2} + 4)}^{2}} =_{x = 2 t} \int_{\frac{1}{2}}^{\infty} \frac{2 dt}{16 {(t^{2} + 1)}^{2}} = \frac{1}{8} \int_{\frac{1}{2}}^{\infty} \frac{dt}{{(1 + t^{2})}^{2}}

=_{t = tanu} \frac{1}{8} \int_{\arctan (\frac{1}{2})}^{\frac{π}{2}} \frac{1 + \tan^{2} u}{{(1 + \tan^{2} u)}^{2}} du

= \frac{1}{8} \int_{\arctan (\frac{1}{2})}^{\frac{π}{2}} \frac{1 + \cos (2 u)}{2} du = \frac{1}{16} \int_{\arctan (\frac{1}{2})}^{\frac{π}{2}} (1 + \cos (2 u) du

= \frac{1}{16} (\frac{π}{2} - \arctan (\frac{1}{2})) + \frac{1}{32} {[\sin (2 u)]}_{\dots}^{\dots}

= \frac{\arctan 2}{16} + \frac{1}{32} (- \sin (2 \arctan (\frac{1}{2})) so the value of I is known