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

Question Number 98105 by mathmax by abdo last updated on 11/Jun/20

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

Answered by MJS last updated on 11/Jun/20

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

[Ostrogradski]

= - \frac{44 x^{2} + 55 x + 8}{686 (x - 2) {(2 x + 3)}^{2}} - \frac{11}{343} \int \frac{d x}{(x - 2) (2 x + 3)} =

= - \frac{44 x^{2} + 55 x + 8}{686 (x - 2) {(2 x + 3)}^{2}} + \frac{11}{2401} \ln ∣ \frac{2 x + 3}{x - 2} ∣ + C

\underset{3}{\int^{+ \infty}} \frac{x + 1}{{(x - 2)}^{2} {(2 x + 3)}^{3}} d x = \frac{569}{55566} + \frac{11}{2401} (\ln 2 - 2 \ln 3)

Commented by mathmax by abdo last updated on 11/Jun/20

thank you sir MJS

Answered by mathmax by abdo last updated on 11/Jun/20

binome method I = \int_{3}^{+ \infty} \frac{(x + 1) dx}{{(x - 2)}^{2} {(2 x + 3)}^{3}} \Rightarrow

I = \int_{3}^{+ \infty} \frac{(x + 1) dx}{{(\frac{x - 2}{2 x + 3})}^{2} {(2 x + 3)}^{5}} changement \frac{x - 2}{2 x + 3} = t give x - 2 = 2 tx + 3 t \Rightarrow

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

= \frac{3 - 6 t + 6 t + 4}{{(1 - 2 t)}^{2}} = \frac{7}{{(1 - 2 t)}^{2}}, 2 x + 3 = \frac{6 t + 4}{1 - 2 t} + 3 = \frac{6 t + 4 + 3 - 6 t}{1 - 2 t} = \frac{7}{1 - 2 t}

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

I = \int_{\frac{1}{9}}^{\frac{1}{2}} \frac{(t + 3)}{(1 - 2 t) t^{2} {(\frac{7}{1 - 2 t})}^{5}} \times \frac{7}{{(1 - 2 t)}^{2}} dt = \frac{1}{7^{6}} \int_{\frac{1}{9}}^{\frac{1}{2}} \frac{(t + 3) {(1 - 2 t)}^{5}}{{(1 - 2 t)}^{3} t^{2}} dt

= \frac{1}{7^{6}} \int_{\frac{1}{9}}^{\frac{1}{2}} \frac{(t + 3) {(1 - 2 t)}^{2}}{t^{2}} dt

= \frac{1}{7^{6}} \int_{\frac{1}{9}}^{\frac{1}{2}} \frac{(t + 3) ({4 t}^{2} - 4 t + 1)}{t^{2}} dt

= \frac{1}{7^{6}} \int_{\frac{1}{9}}^{\frac{1}{9}} \frac{{4 t}^{3} - {4 t}^{2} + t + {12 t}^{2} - 12 t + 3}{t^{2}} dt

= \frac{1}{7^{6}} \int_{\frac{1}{9}}^{\frac{1}{9}} \frac{{4 t}^{3} + {8 t}^{2} - 11 t + 3}{t^{2}} dt

= \frac{1}{7^{6}} \int

Commented by mathmax by abdo last updated on 11/Jun/20

I = \frac{1}{7^{6}} {\int_{\frac{1}{9}}^{\frac{1}{2}} (4 t + 8 - \frac{11}{t} + \frac{3}{t^{2}}) dt}

= \frac{1}{7^{6}} {[{2 t}^{2} + 8 t - 11 \ln ∣ t ∣ - \frac{3}{t}]}_{\frac{1}{9}}^{\frac{1}{2}}

= \frac{1}{7^{6}} (\frac{1}{2} + 4 + 11 \ln (2) - 6 - 2 {(\frac{1}{9})}^{2} - 8 \times \frac{1}{9} 11 \ln (\frac{1}{9}) + 27) = \dots