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Question Number 147683 by mathmax by abdo last updated on 22/Jul/21

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

1) find \int F (x) dx

2) en deduire la decomposition de F en element simples

Answered by mathmax by abdo last updated on 23/Jul/21

1) I = \int \frac{dx}{{(x + 1)}^{5} {(2 x - 3)}^{4}} \Rightarrow I = \int \frac{dx}{{(\frac{x + 1}{2 x - 3})}^{5} {(2 x - 3)}^{9}}

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

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

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

I (x) = \int \frac{- 5}{{(2 t - 1)}^{2} t^{5} {(\frac{5}{2 t - 1})}^{9}} dt

= - \frac{1}{5^{8}} \int \frac{{(2 t - 1)}^{9}}{{(2 t - 1)}^{2} t^{5}} dt = - \frac{1}{5^{8}} \int \frac{{(2 t - 1)}^{7}}{t^{5}} dt

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

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

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

= \frac{1}{5^{8}} \sum_{k = 0 and k \neq 4}^{7} {(- 2)}^{k} C_{7}^{k} \times \frac{1}{k - 4} t^{k - 4} + \frac{1}{5^{8}} {(- 2)}^{4} C_{7}^{4} \log ∣ t ∣ + C

I (x) = \frac{1}{5^{8}} \sum_{k = 0 and k \neq 4}^{7} \frac{{(- 2)}^{k}}{k - 4} C_{7}^{k} {(\frac{x + 1}{2 x - 3})}^{k - 4} + \frac{2^{4}}{5^{8}} C_{7}^{4} \log ∣ \frac{x + 1}{2 x - 3} ∣ + C

Commented by mathmax by abdo last updated on 23/Jul/21

F (x) = \frac{dI}{dx} (x)

= \frac{1}{5^{8}} \sum_{k = 0 and k \neq 4}^{7} {(- 2)}^{k} C_{7}^{k} {(\frac{x + 1}{2 x - 3})}^{^{'}} {(\frac{x + 1}{2 x - 3})}^{k - 5} + \frac{2^{4}}{5^{8}} C_{7}^{4} \frac{{(\frac{x + 1}{2 x - 3})}^{^{'}}}{\frac{x + 1}{2 x - 3}}

= - \frac{1}{5^{7}} \sum_{k = 0 andk \neq 4}^{7} {(- 2)}^{k} C_{7}^{k} \frac{{(x + 1)}^{k - 5}}{{(2 x - 3)}^{k - 3}} - \frac{2^{4}}{5^{7}} C_{7}^{4} \frac{2 x - 3}{x + 1} \times \frac{1}{{(2 x - 3)}^{2}}

\dots . be continued \dots .