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Question Number 135892 by mathmax by abdo last updated on 16/Mar/21

1) decompose inside R (x) the fraction F (x) = \frac{1}{{(x + 2)}^{3} {(x - 1)}^{4}}

2) calculate \int_{2}^{\infty} F (x) dx

3) calculate \int_{2}^{\infty} F^{2} (x) dx

Answered by mathmax by abdo last updated on 17/Mar/21

1) F (x) = \sum_{i = 1}^{3} \frac{a_{i}}{{(x + 2)}^{i}} + \sum_{i = 1}^{4} \frac{b_{i}}{{(x - 1)}^{i}} whst s a_{i} ?

let find D_{3} (- 2) for f (x) = {(x - 1)}^{- 4}

f (x) = f (- 2) + \frac{x + 2}{1!} f^{^{'}} (- 2) + \frac{{(x + 2)}^{2}}{2!} f^{(2)} (- 2) + \frac{{(x + 2)}^{3}}{3!} f^{(3)} (- 2) + \frac{{(x + 2)}^{4}}{4!} ξ (x)

f (- 2) = {(- 3)}^{- 4}, f^{^{'}} (x) = - 4 {(x - 1)}^{- 5} \Rightarrow f^{^{'}} (- 2) = - 4 {(- 3)}^{- 5}

f^{(2)} (x) = 20 {(x - 1)}^{- 6} \Rightarrow f^{(2)} (- 2) = 20 {(- 3)}^{- 6}

f^{(3)} (x) = - 120 {(x - 1)}^{- 7} \Rightarrow f^{(3)} (- 2) = - 120 {(- 3)}^{- 7} \Rightarrow

f (x) = {(- 3)}^{- 4} - 4 {(- 3)}^{- 5} (x + 2) + \frac{20 {(- 3)}^{- 6}}{2} {(x + 2)}^{2} - \frac{120 {(- 3)}^{- 7}}{3!} {(x + 2)}^{3}

+ \frac{{(x + 2)}^{4}}{4!} ξ (x) \Rightarrow

\frac{f (x)}{{(x + 2)}^{3}} = \frac{{(- 3)}^{- 4}}{{(x + 2)}^{3}} - \frac{4 {(- 3)}^{- 5}}{{(x + 2)}^{2}} + \frac{10 {(- 3)}^{- 6}}{(x + 2)} + \dots \Rightarrow a_{1} = 10 . {(- 3)}^{- 6}

a_{2} = - 4 {(- 3)}^{- 5}, a_{3} = {(- 3)}^{- 4} let find b_{i} we find D_{3} (1) for g (x) = {(x + 2)}^{- 3}

g (x) = g (1) + \frac{x - 1}{1!} g^{^{'}} (1) + \frac{{(x - 1)}^{2}}{2!} g^{(2)} (1) + \frac{{(x - 1)}^{3}}{3!} g^{(3)} (1) + \frac{{(x - 1)}^{4}}{4!} δ (x)

g (1) = 3^{- 3}, g^{^{'}} (x) = - 3 {(x + 2)}^{- 4} \Rightarrow g^{^{'}} (1) = - 3 . 3^{- 4}

g^{(2)} (x) = 12 {(x + 2)}^{- 5} \Rightarrow g^{(2)} (1) = 12 . 3^{- 5} g^{(3)} (x) = 12.5 {(x + 2)}^{- 6} \Rightarrow g^{(3)} (1) = 60 . 3^{- 6}

g (x) = 3^{- 3} - 3 . 3^{- 4} (x - 1) + \frac{12 . 3^{- 5}}{2} {(x - 1)}^{2} + \frac{60 . 3^{- 6}}{3!} {(x - 1)}^{3} + \dots

= \frac{1}{27} - \frac{1}{27} (x - 1) + \frac{2}{81} {(x - 1)}^{2} + 10 . 3^{- 6} {(x - 1)}^{3} + \frac{{(x - 1)}^{4}}{4!} δ (x)

\Rightarrow \frac{g (x)}{{(x - 1)}^{4}} = \frac{1}{27 {(x - 1)}^{4}} - \frac{1}{27 {(x - 1)}^{3}} + \frac{2}{81 {(x - 1)}^{2}} + \frac{10 . 3^{- 6}}{x - 1} \Rightarrow

b_{1} = 10 . 3^{- 6}, b_{2} = \frac{2}{81}, b_{3} = - \frac{1}{27}, b_{4} = \frac{1}{27} \Rightarrow

Answered by mathmax by abdo last updated on 17/Mar/21

2) \int_{2}^{\infty} \frac{dx}{{(x + 2)}^{3} {(x - 1)}^{4}} = \int_{2}^{\infty} \frac{dx}{{(\frac{x - 1}{x + 2})}^{4} {(x + 2)}^{7}} we do the changement

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

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

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

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

= \frac{1}{3^{6}} \int_{\frac{1}{4}}^{1} \frac{{(1 - t)}^{5}}{t^{4}} dt = - \frac{1}{3^{6}} \int_{\frac{1}{4}}^{1} \frac{\sum_{k = 0}^{5} C_{5}^{k} t^{k} {(- 1)}^{5 - k}}{t^{4}} dt

= \frac{1}{3^{6}} \sum_{k = 0}^{5} {(- 1)}^{k} C_{5}^{k} \int_{\frac{1}{4}}^{1} t^{k - 4} dt

= \frac{1}{3^{6}} \sum_{k = 0 and k \neq 3}^{5} {(- 1)}^{k} C_{5}^{k} {[\frac{1}{k - 3} t^{k - 3}]}_{\frac{1}{4}}^{1} - \frac{1}{3^{6}} C_{5}^{3} {[\ln ∣ t ∣]}_{\frac{1}{4}}^{1}

= \frac{1}{3^{6}} \sum_{k = 0 and k \neq 3}^{5} \frac{{(- 1)}^{k} C_{5}^{k}}{k - 3} (1 - \frac{1}{4^{k - 3}}) + \frac{1}{3^{6}} C_{5}^{3} (2 \ln (2))

Commented by mathmax by abdo last updated on 17/Mar/21

let Φ = \int_{2}^{\infty} F^{2} (x) dx \Rightarrow Φ = \int_{2}^{\infty} \frac{dx}{{(x + 2)}^{6} {(x - 1)}^{8}}

= \int_{2}^{\infty} \frac{dx}{{(\frac{x - 1}{x + 2})}^{8} {(x + 2)}^{14}} chamgement \frac{x - 1}{x + 2} = t give

Φ = \int_{\frac{1}{4}}^{1} \frac{1}{t^{8} {(\frac{3}{1 - t})}^{14}} \frac{3 dt}{{(1 - t)}^{2}} = \frac{1}{3^{13}} \int_{\frac{1}{4}}^{1} \frac{{(1 - t)}^{12}}{t^{8}} dt

= \frac{1}{3^{13}} \int_{\frac{1}{4}}^{1} \frac{1}{t^{8}} \sum_{k = 0}^{12} C_{12}^{k} t^{k} {(- 1)}^{12 - k} dt

= \frac{1}{3^{13}} \sum_{k = 0}^{12} {(- 1)}^{k} C_{12}^{k} \int_{\frac{1}{4}}^{1} t^{k - 8} dt

= \frac{1}{3^{13}} \sum_{k = 0 and k \neq 7}^{12} {(- 1)}^{k} C_{12}^{k} {[\frac{1}{k - 7} t^{k - 7}]}_{\frac{1}{4}}^{1} - \frac{1}{3^{13}} C_{12}^{7} {[\ln ∣ t ∣]}_{\frac{1}{4}}^{1}

Φ = \frac{1}{3^{13}} \sum_{k = 0 and k \neq 7}^{12} \frac{{(- 1)}^{k} C_{12}^{k}}{k - 7} (1 - \frac{1}{4^{k - 7}}) + \frac{1}{3^{13}} C_{12}^{7} (2 \ln (2))