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

Question Number 103591 by mathmax by abdo last updated on 16/Jul/20

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

Commented by Worm_Tail last updated on 16/Jul/20

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

\int_{3}^{o o} \frac{- 3}{16 (x + 1)} + \frac{4}{27 (x + 2)} + \frac{1}{16 {(x + 1)}^{2}} + \frac{1}{27 {(x + 2)}^{2}} - \frac{1}{8 {(x + 1)}^{3}} + \frac{17}{432 (x - 1)} - \frac{13}{432 {(x - 1)}^{2}} + \frac{1}{72 {(x - 1)}^{3}} d x

{[\frac{- 3 l n (x + 1)}{16} + \frac{4}{27} l n (x + 2) - \frac{1}{16 (x + 1)} - \frac{1}{27 (x + 2)} + \frac{1}{16 {(x + 1)}^{2}} + \frac{17 l n (x - 1)}{432} + \frac{13}{432 (x - 1)} - \frac{1}{144 {(x - 1)}^{2}}]}_{3}^{o o}

{[\frac{l n \frac{{(x + 2)}^{64} {(x - 1)}^{17}}{{(x + 1)}^{81}}}{432} - \frac{1}{27 (x + 2)} + \frac{1}{16 {(x + 1)}^{2}} + \frac{13}{432 (x - 1)} - \frac{1}{144 {(x - 1)}^{2}}]}_{3}^{o o}

l i \underset{t \to o o}{m} [\frac{l n \frac{{(t + 2)}^{64} {(t - 1)}^{17}}{{(t + 1)}^{81}}}{432} - \frac{1}{27 (t + 2)} + \frac{1}{16 {(t + 1)}^{2}} + \frac{13}{432 (t - 1)} - \frac{1}{144 {(t - 1)}^{2}}]

- [\frac{l n \frac{{(3 + 2)}^{64} {(3 - 1)}^{17}}{{(3 + 1)}^{81}}}{432} - \frac{1}{27 (3 + 2)} + \frac{1}{16 {(3 + 1)}^{2}} + \frac{13}{432 (3 - 1)} - \frac{1}{144 {(3 - 1)}^{2}}]

l i \underset{t \to o o}{m} [\frac{l n (1)}{432} - 0 + 0 + 0 - 0] - [\frac{l n \frac{5^{64} \times 2^{17}}{4^{81}}}{432} + \frac{113}{11520}]

- [\frac{64 l n (5) - 145 l n (2)}{432} + \frac{113}{11520}]

\frac{- 64 l n (5) + 145 l n (2)}{432} - \frac{113}{11520}

Commented by Worm_Tail last updated on 16/Jul/20

m i s t a k e m i g h t h a v e c r e p t i n

Commented by abdomathmax last updated on 16/Jul/20

can you chow how do you find the decomposition ?

Commented by Worm_Tail last updated on 16/Jul/20

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

\frac{A}{(x + 1)} + \frac{B}{(x + 2)} + \frac{C}{{(x + 1)}^{2}} + \frac{D}{{(x + 2)}^{2}} + \frac{E}{{(x + 1)}^{3}} + \frac{F}{(x - 1)} + \frac{G}{{(x - 1)}^{2}} + \frac{H}{{(x - 1)}^{3}} = \frac{1}{{(x^{2} - 1)}^{3} {(x + 2)}^{2}}

y o u c o n t i n u e f r o m h e r e

Answered by abdomathmax last updated on 16/Jul/20

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

= \int_{3}^{\infty} \frac{dx}{{(\frac{x - 1}{x + 1})}^{3} {(x + 1)}^{6} {(x + 2)}^{2}} we do tbe cha 7 gement

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

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

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

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

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

= 2 \int_{\frac{1}{2}}^{1} \frac{{(t - 1)}^{8}}{{(t - 1)}^{2} t^{3} . 2^{6} {(3 - t)}^{2}} dt

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

again ch . \frac{t}{t - 3} = z \Rightarrow t = zt - 3 z \Rightarrow (1 - z) t = - 3 z \Rightarrow

t = \frac{- 3 z}{1 - z} = \frac{3 z}{z - 1} and t - 3 = \frac{3 z}{z - 1} - 3 = \frac{3 z - 3 z + 3}{z - 1}

= \frac{3}{z - 1} and t - 1 = \frac{3 z}{z - 1} - 1 = \frac{3 z - z + 1}{z - 1} = \frac{2 z + 1}{z - 1}

I = \frac{1}{2^{5}} \int_{- \frac{1}{5}}^{- \frac{1}{2}} \frac{{(\frac{2 z + 1}{z - 1})}^{6}}{z^{3} {(\frac{3}{z - 1})}^{5}} \times \frac{- 3}{{(z - 1)}^{2}} dz

= - \frac{1}{2^{5} . 3^{4}} \int_{- \frac{1}{5}}^{- \frac{1}{2}} \frac{{(2 z + 1)}^{6} {(z - 1)}^{5}}{z^{3} {(z - 1)}^{8}} dz

= - \frac{1}{2^{5} . 3^{4}} \int_{- \frac{1}{5}}^{- \frac{1}{2}} \frac{{(2 z + 1)}^{6}}{z^{3} {(z - 1)}^{3}} dz

= - \frac{1}{2^{5} . 3^{4}} \int_{- \frac{1}{5}}^{- \frac{1}{2}} \frac{\sum_{k = 0}^{6} C_{6}^{k} {(2 z)}^{k}}{z^{3} {(z - 1)}^{3}} dz

= - \frac{1}{2^{5} . 3^{4}} \int_{- \frac{1}{5}}^{- \frac{1}{2}} \sum_{k = 0}^{6} 2^{k} C_{6}^{k} \frac{z^{k}}{z^{3} {(z - 1)}^{3}} dz

= - \frac{1}{2^{5} . 3^{4}} \sum_{k = 0}^{6} 2^{k} C_{6}^{k} \int_{- \frac{1}{5}}^{- \frac{1}{2}} \frac{z^{k}}{z^{3} {(z - 1)}^{3}} dz

after we decompose F_{k} (z) = \frac{z^{k}}{z^{3} {(z - 1)}^{3}}

\dots be continued \dots