1-find-dx-x-1-2-x-3-4-2-deduce-the-decomposition-of-F-x-1-x-1-2-x-3-4-

Question Number 135957 by mathmax by abdo last updated on 17/Mar/21

1) find \int \frac{dx}{{(x + 1)}^{2} {(x - 3)}^{4}}

2) deduce the decomposition of F (x) = \frac{1}{{(x + 1)}^{2} {(x - 3)}^{4}}

Answered by Dwaipayan Shikari last updated on 17/Mar/21

\frac{1}{{(x + 1)}^{2} {(x - 3)}^{4}} = \frac{1}{{(x - 3)}^{2}} {(\frac{1}{(x + 1) (x - 3)})}^{2} = τ (x)

= \frac{1}{16} (\frac{1}{{(x - 3)}^{2} {(x + 1)}^{2}} - \frac{2}{{(x - 3)}^{3} (x + 1)} + \frac{1}{{(x - 3)}^{4}})

= \frac{1}{256} (\frac{1}{{(x - 3)}^{2}} + \frac{1}{{(x + 1)}^{2}} - \frac{2}{(x - 3) (x + 1)}) - \frac{1}{32 {(x - 3)}^{2}} (\frac{1}{x - 3} - \frac{1}{x + 1}) + \frac{1}{16 {(x - 3)}^{4}}

= \frac{1}{256 {(x - 3)}^{2}} + \frac{1}{256 {(x + 1)}^{2}} - \frac{1}{512 (x - 3)} + \frac{1}{512 (x + 1)} - \frac{1}{32 {(x - 3)}^{3}} + \frac{1}{128 {(x - 3)}^{2}} - \frac{1}{512 (x - 3)} + \frac{1}{512 (x + 1)} + \frac{1}{16 {(x - 3)}^{4}}

= \frac{3}{256 {(x - 3)}^{2}} - \frac{1}{256 (x - 3)} + \frac{1}{256 (x + 1)} + \frac{1}{256 {(x + 1)}^{2}} + \frac{1}{32 {(x - 3)}^{3}} + \frac{1}{16 {(x - 3)}^{4}}

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

1) Φ = \int \frac{dx}{{(x + 1)}^{2} {(x - 3)}^{4}} \Rightarrow Φ = \int \frac{dx}{{(\frac{x - 3}{x + 1})}^{4} {(x + 1)}^{6}}

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

x = \frac{3 + t}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{1 - t - (3 + t) (- 1)}{{(1 - t)}^{2}} = \frac{1 - t + 3 + t}{{(1 - t)}^{2}} = \frac{4}{{(1 - t)}^{2}} and

x + 1 = \frac{3 + t}{1 - t} + 1 = \frac{3 + t + 1 - t}{1 - t} = \frac{4}{1 - t} \Rightarrow

Φ = \int \frac{1}{t^{4} {(\frac{4}{1 - t})}^{6}} \frac{4 dt}{{(1 - t)}^{2}} = \frac{1}{4^{5}} \int \frac{{(1 - t)}^{4}}{t^{4}} dt

= \frac{1}{4^{5}} \int \frac{{(t - 1)}^{4}}{t^{4}} dt = \frac{1}{4^{5}} \int \frac{{(t^{2} - 2 t + 1)}^{2}}{t^{4}} dt

= \frac{1}{4^{5}} \int \frac{{(t^{2} - 2 t)}^{2} + 2 (t^{2} - 2 t) + 1}{t^{4}} dt

= \frac{1}{4^{5}} \int \frac{t^{4} - {4 t}^{3} + {4 t}^{2} + {2 t}^{2} - 4 t + 1}{t^{4}} dt

= \frac{1}{4^{5}} \int \frac{t^{4} - {4 t}^{3} + {6 t}^{2} - 4 t + 1}{t^{4}} dt

= \frac{1}{4^{5}} {\int (1 - \frac{4}{t} + \frac{6}{t^{2}} - \frac{4}{t^{3}} + \frac{1}{t^{4}}) dt}

\Rightarrow 4^{5} . Φ = t - 4 \ln ∣ t ∣ - \frac{6}{t} - 4 . \frac{1}{- 3 + 1} t^{- 3 + 1} + \frac{1}{- 4 + 1} t^{- 4 + 1} + C

= t - 4 \ln ∣ t ∣ - \frac{6}{t} + \frac{2}{t^{2}} - \frac{1}{{3 t}^{3}} + C

= \frac{x - 3}{x + 1} - 4 \ln ∣ \frac{x - 3}{x + 1} ∣ - 6 . \frac{x + 1}{x - 3} + 2 {(\frac{x + 1}{x - 3})}^{2} - \frac{1}{3} {(\frac{x + 1}{x - 3})}^{3} + C \Rightarrow

Φ = \frac{1}{4^{5}} {\frac{x - 3}{x + 1} - 4 \ln ∣ \frac{x - 3}{x + 1} ∣ - \frac{6 (x + 1)}{x - 3} + 2 {(\frac{x + 1}{x - 3})}^{2} - \frac{1}{3} {(\frac{x + 1}{x - 3})}^{2}} + C

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

2) F (x) = \frac{d}{dx} Φ

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

= {(\ln ∣ \frac{x - 3}{x + 1} ∣)}^{(1)} = \frac{4}{{(x + 1)}^{2}} \times \frac{x + 1}{x - 3} = \frac{4}{(x + 1) (x - 3)} = \frac{1}{x - 3} - \frac{1}{x + 1}

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

{{(\frac{x + 1}{x - 3})}^{2}}^{(1)} = 2 (\frac{x + 1}{x - 3}) \times \frac{- 4}{{(x - 3)}^{2}} = - \frac{8 (x + 1)}{{(x - 3)}^{3}}

= - 8 \frac{x - 3 + 4}{{(x - 3)}^{3}} = - \frac{8}{{(x - 3)}^{2}} - \frac{32}{{(x - 3)}^{3}}

{(\frac{x + 1}{x - 3})}^{3}}^{(1)} = 3 {(\frac{x + 1}{x - 3})}^{2} . \frac{- 4}{{(x - 3)}^{2}} = - 12 \frac{{(x + 1)}^{2}}{{(x - 3)}^{3}}

= - 12 \frac{{(x - 3 + 4)}^{2}}{{(x - 3)}^{3}} = - 12 . \frac{{(x - 3)}^{2} + 8 (x - 3) + 16}{{(x - 3)}^{3}}

= - 12 {\frac{1}{(x - 3)} + \frac{8}{{(x - 3)}^{2}} + \frac{16}{{(x - 3)}^{3}}} rest to collect the values \dots