calculate-0-dx-2x-1-4-x-3-5-

Question Number 146196 by mathmax by abdo last updated on 11/Jul/21

calculate \int_{0}^{\infty} \frac{dx}{{(2 x + 1)}^{4} {(x + 3)}^{5}}

Answered by mathmax by abdo last updated on 12/Jul/21

Ψ = \int_{0}^{\infty} \frac{dx}{{(2 x + 1)}^{4} {(x + 3)}^{5}} \Rightarrow Ψ = \int_{0}^{\infty} \frac{dx}{{(\frac{2 x + 1}{x + 3})}^{4} {(x + 3)}^{9}}

we do the changement \frac{2 x + 1}{x + 3} = t \Rightarrow 2 x + 1 = tx + 3 t \Rightarrow (2 - t) x = 3 t - 1

\Rightarrow x = \frac{3 t - 1}{2 - t} \Rightarrow \frac{dx}{dt} = \frac{3 (2 - t) - (3 t - 1) (- 1)}{{(2 - t)}^{2}} = \frac{6 - 3 t + 3 t - 1}{{(2 - t)}^{2}} = \frac{5}{{(2 - t)}^{2}}

x + 3 = \frac{3 t - 1}{2 - t} + 3 = \frac{3 t - 1 + 6 - 3 t}{2 - t} = \frac{5}{2 - t} \Rightarrow

Ψ = \int_{\frac{1}{3}}^{2} \frac{5}{{(2 - t)}^{2} t^{4} {(\frac{5}{2 - t})}^{9}} dt = \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{{(2 - t)}^{7}}{t^{4}} dt

= \frac{1}{5^{8}} \int_{\frac{1}{3}}^{2} \frac{\sum_{k = 0}^{7} C_{7}^{k} {(- t)}^{k} 2^{7 - k}}{t^{4}} dt

= \frac{2^{7}}{5^{8}} \int_{\frac{1}{3}}^{2} \sum_{k = 0}^{7} C_{7}^{k} {(- 1)}^{k} t^{k - 4} 2^{- k} dt

= \frac{2^{7}}{5^{8}} \sum_{k = 0}^{7} {(- 1)}^{k} 2^{- k} C_{7}^{k} \int_{\frac{1}{3}}^{2} t^{k - 4} dt

= \frac{2^{7}}{5^{8}} {\sum_{k = 0 and k \neq 3}^{7} {(- \frac{1}{2})}^{k} C_{7}^{k} {[\frac{1}{k - 3} t^{k - 3}]}_{\frac{1}{3}}^{2} - \frac{1}{8} C_{7}^{3} {[\log ∣ t ∣]}_{\frac{1}{3}}^{2}}

Ψ = \frac{2^{7}}{5^{8}} \sum_{k = 0 and k \neq 3}^{7} {(- \frac{1}{2})}^{k} \frac{C_{7}^{k}}{k - 3} (2^{k - 3} - \frac{1}{3^{k - 3}})

- \frac{2^{7}}{8 . 5^{8}} C_{7}^{3} (\log 2 + \log 3)

Answered by Olaf_Thorendsen last updated on 12/Jul/21

R (x) = \frac{1}{{(2 x + 1)}^{4} {(x + 3)}^{5}}

R (x) = \frac{\frac{32}{3125}}{{(2 x + 1)}^{4}} - \frac{\frac{32}{3125}}{{(2 x + 1)}^{3}} + \frac{\frac{96}{15625}}{{(2 x + 1)}^{2}}

- \frac{\frac{224}{78125}}{2 x + 1} + \frac{\frac{1}{625}}{{(x + 3)}^{5}} + \frac{\frac{8}{3125}}{{(x + 3)}^{4}} + \frac{\frac{8}{3125}}{{(x + 3)}^{3}}

+ \frac{\frac{32}{15625}}{{(x + 2)}^{2}} + \frac{\frac{112}{78125}}{x + 3}

\int R (x) d x = - \frac{\frac{16}{9375}}{{(2 x + 1)}^{3}} + \frac{\frac{8}{3125}}{{(2 x + 1)}^{2}}

- \frac{\frac{48}{15625}}{2 x + 1} - \frac{112}{78125} \ln (2 x + 1) - \frac{\frac{1}{2500}}{{(x + 3)}^{4}}

- \frac{\frac{8}{9375}}{{(x + 3)}^{3}} - \frac{\frac{4}{3125}}{{(x + 3)}^{2}} - \frac{\frac{32}{15625}}{x + 3}

+ \frac{112}{78125} \ln (x + 3)

Ω = \int_{0}^{\infty} R (x) d x = \frac{15593}{5062500} - \frac{112}{78125} \ln 6

Commented by mathmax by abdo last updated on 12/Jul/21

thank you sir .