calculate-5-dx-x-2-9-4-

Question Number 104197 by mathmax by abdo last updated on 20/Jul/20

calculate \int_{5}^{+ \infty} \frac{dx}{{(x^{2} - 9)}^{4}}

Answered by Dwaipayan Shikari last updated on 20/Jul/20

\int_{5}^{+ \infty} \frac{2 xdx}{2 x {(x^{2} - 9)}^{4}} = \frac{1}{2} \int_{16}^{+ \infty} \frac{dt}{\sqrt{9 + t}} . \frac{1}{t^{4}} {t = {9 \tan}^{2} θ, 1 = 18 \tan θ \sec^{2} θ \frac{d θ}{dt}

\frac{1}{6} \int_{\tan^{- 1} \frac{4}{3}}^{\frac{π}{2}} \frac{18 \tan θ \sec^{2} θ d θ}{\sec θ . 9^{4} \tan^{8} θ} {take x^{2} - 9 = t

\frac{1}{3} . \frac{1}{9^{3}} \int_{\tan^{- 1} \frac{4}{3}}^{\frac{π}{2}} \frac{\sec θ}{\tan^{7} θ} d θ \dots . . continue

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

A = \int_{5}^{\infty} \frac{dx}{{(x^{2} - 9)}^{4}} \Rightarrow A = \int_{5}^{\infty} \frac{dx}{{(x - 3)}^{4} {(x + 3)}^{4}} = \int_{5}^{\infty} \frac{dx}{{(\frac{x - 3}{x + 3})}^{4} {(x + 3)}^{8}}

we do the changement \frac{x - 3}{x + 3} = t \Rightarrow x - 3 = tx + 3 t \Rightarrow (1 - t) x = 3 t + 3 \Rightarrow x = \frac{3 t + 3}{1 - t}

\frac{dx}{dx} = \frac{3 (1 - t) + (3 t + 3)}{{(1 - t)}^{2}} = \frac{6}{{(1 - t)}^{2}}

x + 3 = \frac{3 t + 3}{1 - t} + 3 = \frac{3 t + 3 + 3 - 3 t}{1 - t} = \frac{6}{1 - t} \Rightarrow A = \int_{\frac{1}{4}}^{1} \frac{6}{{(1 - t)}^{2} t^{4} {(\frac{6}{1 - t})}^{8}} dt

= \frac{1}{6^{7}} \int_{\frac{1}{4}}^{1} \frac{{(t - 1)}^{8}}{{(t - 1)}^{2} t^{4}} dt = \frac{1}{6^{7}} \int_{\frac{1}{4}}^{1} \frac{{(t - 1)}^{6}}{t^{4}} dt

= \frac{1}{6^{7}} \int_{\frac{1}{4}}^{1} \frac{\sum_{k = 0}^{6} C_{6}^{k} t^{k}}{t^{4}} dt = \frac{1}{6^{7}} \sum_{k = 0}^{6} C_{6}^{k} \int_{\frac{1}{4}}^{1} t^{k - 4} dt

= \frac{1}{6^{7}} \sum_{k = 0 and k \neq 3}^{6} C_{6}^{k} {[\frac{1}{k - 3} t^{k - 3}]}_{\frac{1}{4}}^{1} + \frac{1}{6^{7}} C_{6}^{3} {[lnt]}_{\frac{1}{4}}^{1}

A = \frac{1}{6^{7}} \sum_{k = 0 and k \neq 3}^{6} \frac{C_{6}^{k}}{k - 3} {1 - \frac{1}{4^{k - 3}}} + \frac{C_{6}^{3}}{6^{7}} \times (2 \ln 2)