calculate-1-dx-x-3-2x-1-4-1-without-use-of-decomposition-2-by-use-of-decomposition-

Question Number 95584 by turbo msup by abdo last updated on 26/May/20

c a l c u l a t e \int_{1}^{+ \infty} \frac{d x}{x^{3} {(2 x + 1)}^{4}}

1) w i t h o u t u s e o f d e c o m p o s i t i o n

2) b y u s e o f d e c o m p o s i t i o n

Answered by MJS last updated on 26/May/20

Ostrogradski leads to

\frac{960 x^{4} + 1200 x^{3} + 440 x^{2} + 30 x - 3}{6 x^{2} {(2 x + 1)}^{3}} + 40 \int \frac{d x}{x (2 x + 1)} =

= \frac{960 x^{4} + 1200 x^{3} + 440 x^{2} + 30 x - 3}{6 x^{2} {(2 x + 1)}^{3}} + 40 \ln \frac{x}{2 x + 1} + C

\Rightarrow

\underset{1}{\int^{+ \infty}} \frac{d x}{x^{3} {(2 x + 1)}^{4}} = 40 \ln \frac{3}{2} - \frac{2627}{162}

Commented by mathmax by abdo last updated on 26/May/20

thank you sir mjs

Answered by mathmax by abdo last updated on 26/May/20

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

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

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

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

= - \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{{(2 t - 1)}^{5}}{t^{3}} dt = - \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\sum_{k = 0}^{5} C_{5}^{k} {(2 t)}^{k} {(- 1)}^{5 - k}}{t^{3}} dt

= \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\sum_{k = 0}^{5} 2^{k} {(- 1)}^{k} C_{5}^{k} t^{k}}{t^{3}} dt

= \sum_{k = 0}^{5} C_{5}^{k} {(- 2)}^{k} \int_{\frac{1}{3}}^{\frac{1}{2}} t^{k - 3} dt

= \sum_{k = 0 and k \neq 2}^{5} {(- 2)}^{k} C_{5}^{k} {[\frac{1}{k - 2} t^{k - 2}]}_{\frac{1}{3}}^{\frac{1}{2}} + 4 C_{5}^{2} {[lnt]}_{\frac{1}{3}}^{\frac{1}{2}}

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

Answered by mathmax by abdo last updated on 27/May/20

thank you sir mjs

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

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

a_{i} ? we find D_{2} (0) for f (x) = {(x + \frac{1}{2})}^{- 4}

f (x) = f (0) + \frac{x}{1!} f^{^{'}} (0) + \frac{x^{2}}{2!} f^{(2)} (0) + \frac{x^{3}}{3!} ξ (x)

f (0) = \frac{1}{2^{- 4}}, f^{^{'}} (x) = - 4 {(x + \frac{1}{2})}^{- 5} \Rightarrow f^{^{'}} (0) = \frac{- 4}{2^{- 5}}

f^{(2)} (x) = 20 {(x + \frac{1}{2})}^{- 6} \Rightarrow f^{(2)} (0) = \frac{20}{2^{- 6}} \Rightarrow

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

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

a_{2} = - 4 . 2^{5} and a_{3} = 2^{4} let find b_{i}

we determine D_{3} (- \frac{1}{2}) for g (x) = x^{- 3}

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

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

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

g^{(2)} (x) = 12 x^{- 5} \Rightarrow g^{(2)} (- \frac{1}{2}) = 12 {(- \frac{1}{2})}^{- 5}

g^{(3)} (x) = - 60 x^{- 6} \Rightarrow g^{(3)} (- \frac{1}{2}) = - 60 {(- \frac{1}{2})}^{- 6}

\Rightarrow g (x) = {(- \frac{1}{2})}^{- 3} - 3 {(- \frac{1}{2})}^{- 4} (x + \frac{1}{2}) + 6 {(- \frac{1}{2})}^{- 5} {(x + \frac{1}{2})}^{2}

- 10 {(- \frac{1}{2})}^{- 6} {(x + \frac{1}{2})}^{3} + \frac{{(x + \frac{1}{2})}^{4}}{4!} ξ (x)

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

+ \frac{{(x + \frac{1}{2})}^{4}}{4!} ξ (x) \Rightarrow

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

\Rightarrow b_{1} = - 10 . 2^{6}, b_{2} = - 6 . 2^{5}, b_{3} = - 3 . 2^{4}, b_{4} = - 2^{3} \Rightarrow

2^{4} F (x) = \frac{10 . 2^{6}}{x} - \frac{4 . 2^{5}}{x^{2}} + \frac{2^{4}}{x^{3}} - \frac{10 . 2^{6}}{(x + \frac{1}{2})} - \frac{6 . 2^{5}}{{(x + \frac{1}{2})}^{2}} - \frac{3 . 2^{4}}{{(x + \frac{1}{2})}^{3}}

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

F (x) = \frac{10 . 2^{2}}{x} - \frac{4.2}{x^{2}} + \frac{1}{x^{3}} - \frac{10 . 2^{2}}{(x + \frac{1}{2})} - \frac{6.2}{{(x + \frac{1}{2})}^{2}} - \frac{3}{{(x + \frac{1}{2})}^{3}}

- \frac{2^{- 1}}{{(x + \frac{1}{2})}^{4}} now its eazy to find \int_{1}^{+ \infty} F (x) dx \dots