calculate-2-dt-2t-3-4-t-1-5-

Question Number 114056 by mathmax by abdo last updated on 16/Sep/20

calculate \int_{2}^{+ \infty} \frac{dt}{{(2 t + 3)}^{4} {(t - 1)}^{5}}

Answered by Olaf last updated on 17/Sep/20

R (x) =

\frac{\frac{1}{625}}{{(x - 1)}^{5}}

- \frac{\frac{8}{3125}}{{(x - 1)}^{4}}

+ \frac{\frac{8}{3135}}{{(x - 1)}^{3}}

+ \frac{\frac{32}{15625}}{{(x - 1)}^{2}}

+ \frac{\frac{112}{78125}}{x - 1}

+ \frac{\frac{32}{3125}}{{(2 x + 3)}^{4}}

- \frac{\frac{32}{3125}}{{(2 x + 3)}^{3}}

- \frac{\frac{96}{15625}}{{(2 x + 3)}^{2}}

- \frac{\frac{225}{78125}}{2 x + 3}

\int_{2}^{\infty} R (x) d x = \dots

Answered by Olaf last updated on 17/Sep/20

u = t - 1

I = \int_{1}^{\infty} \frac{d u}{u^{5} {(u + 5)}^{4}}

x = \frac{1}{u}

I = \int_{1}^{0} \frac{x^{5}}{{(\frac{1}{x} + 5)}^{5}} (- \frac{d x}{x^{2}})

I = \int_{0}^{1} \frac{x^{8}}{{(5 x + 1)}^{5}} d x

λ = 5 x + 1

I = \frac{1}{5^{8}} \int_{1}^{6} \frac{{(λ - 1)}^{8}}{λ^{5}} . \frac{d λ}{5}

I = \frac{1}{5^{9}} \int_{1}^{6} \frac{\underset{k = 0}{\sum^{8}} C_{8}^{k} λ^{k} {(- 1)}^{8 - k}}{λ^{5}} d λ

\dots m a y b e t h i s w a y i s b e t t e r \dots

Answered by 1549442205PVT last updated on 17/Sep/20

Put u = \frac{1}{t - 1} \Rightarrow du = \frac{- 1}{{(t - 1)}^{2}} dt

dt = - {(t - 1)}^{2} du = \frac{- 1}{u^{2}} du, t = \frac{1}{u} + 1

\int_{2}^{\infty} \frac{dt}{{(2 t + 3)}^{4} {(t - 1)}^{5}} = \int_{1}^{0} \frac{- u^{3} du}{{(\frac{2}{u} + 5)}^{4}}

= - \int_{1}^{0} \frac{u^{7}}{{(5 u + 2)}^{4}} du = - \frac{1}{625} \int_{7 / 5}^{0} \frac{u^{7}}{{(u + \frac{2}{5})}^{4}} du

Put u + \frac{2}{5} = v \Rightarrow du = dv, u = v - \frac{2}{5}

I = - \frac{1}{625} \int_{7 / 5}^{2 / 5} \frac{{(v - \frac{2}{5})}^{7}}{v^{4}} dv

= \frac{- 1}{625} \int_{7 / 5}^{2 / 5} {[v^{7} - {7 v}^{6} . 0.4 + {21 v}^{5} . 0 . 4^{2}

- {35 v}^{4} . 0 . 4^{3} + {35 v}^{3} . 0 . 4^{4} - {21 v}^{2} . {(0.4)}^{5}

+ 7 v . {(0.4)}^{6} - {(0.4)}^{7}] / v^{4}} dv

= \frac{- 1}{625} \int_{7 / 5}^{\infty} [v^{3} - \frac{14}{5} v^{2} + 21 v . {(\frac{2}{5})}^{2} - \frac{35.8}{125}

+ \frac{35.16}{625 v} - \frac{21.32}{{3125 v}^{2}} + \frac{7.64}{{15625 v}^{3}} - \frac{128}{5^{7} v^{4}}]

= \frac{- 1}{625} {\frac{v^{4}}{4} - \frac{{14 v}^{3}}{15} + \frac{42}{25} v^{2} - \frac{280}{125} v

{+ \frac{560}{625} . \ln ∣ v ∣ + \frac{672}{3125 v} - \frac{224}{{15625 v}^{2}} + \frac{128}{3 . 5^{7} v^{3}}]}_{7 / 5}^{2 / 5}

= - \frac{1}{625} [(- 1.0449965 - (- 0.99590280)] =

(- 0.0490937) / 625 = 0.00007855

Answered by MJS_new last updated on 17/Sep/20

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

[{Ostrogradski}^{'} s Method]

= \frac{5376 t^{6} + 1344 t^{5} - 25760 t^{4} + 3080 t^{3} + 40740 t^{2} - 17906 t - 16249}{187500 {(t - 1)}^{4} {(2 t + 3)}^{3}} +

+ \frac{112}{15625} \int \frac{d t}{(t - 1) (2 t + 3)}

this integral = \frac{112}{78125} \ln ∣ \frac{t - 1}{2 t + 3} ∣

\Rightarrow answer is \frac{112}{78125} (\ln 7 - \ln 2) - \frac{36817}{21437500}

Commented by 1549442205PVT last updated on 17/Sep/20

Great! {Sir}^{'} s result coincde to my result

Commented by MJS_new last updated on 17/Sep/20

yes .

I like {Ostrogradski}^{'} s Method better than

decomposing \dots

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