calculate-1-dx-x-2-x-1-3-x-2-4-

Question Number 94906 by mathmax by abdo last updated on 21/May/20

calculate \int_{1}^{+ \infty} \frac{dx}{x^{2} {(x + 1)}^{3} {(x + 2)}^{4}}

Answered by MJS last updated on 22/May/20

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

[Ostrogradski]

= \frac{105 x^{5} + 675 x^{4} + 1565 x^{3} + 1525 x^{2} + 506 x - 12}{24 x {(x + 1)}^{2} {(x + 2)}^{3}} +

+ \frac{5}{8} \int \frac{7 x - 1}{x (x + 1) (x + 2)} d x

\frac{5}{8} \int \frac{7 x - 1}{x (x + 1) (x + 2)} d x =

= - \frac{5}{16} \int \frac{d x}{x} + 5 \int \frac{d x}{x + 1} - \frac{75}{16} \int \frac{d x}{x + 2} =

= - \frac{5}{16} \ln x + 5 \ln (x + 1) - \frac{75}{16} \ln (x + 2) + C

\underset{1}{\int^{\infty}} \frac{d x}{x^{2} {(x + 1)}^{3} {(x + 2)}^{4}} = - \frac{1091}{648} + \frac{75}{16} \ln 3 - 5 \ln 2

Commented by mathmax by abdo last updated on 22/May/20

thank you sir .

Answered by mathmax by abdo last updated on 22/May/20

I = \int_{1}^{+ \infty} \frac{dx}{x^{2} {(x + 1)}^{3} {(x + 2)}^{4}} \Rightarrow I = \int_{1}^{+ \infty} \frac{dx}{{(\frac{x}{x + 1})}^{2} {(x + 1)}^{5} {(x + 2)}^{4}}

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

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

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

I = \int_{\frac{1}{2}}^{1} \frac{dt}{{(t - 1)}^{2} {(\frac{1}{1 - t})}^{5} {(\frac{t - 2}{t - 1})}^{4}} = - \int_{\frac{1}{2}}^{1} \frac{{(t - 1)}^{9}}{{(t - 1)}^{2} {(t - 2)}^{4}} dt

= - \int_{\frac{1}{2}}^{1} \frac{{(t - 1)}^{7}}{{(t - 2)}^{4}} dt =_{t - 2 = z} - \int_{- \frac{3}{2}}^{- 1} \frac{{(z + 1)}^{7}}{z^{4}} dz

= - \int_{- \frac{3}{2}}^{- 1} \frac{\sum_{k = 0}^{7} C_{7}^{k} z^{k}}{z^{4}} dz = - \int_{- \frac{3}{2}}^{- 1} \sum_{k = 0}^{7} C_{7}^{k} z^{k - 4} dz

= - \sum_{k = 0}^{7} C_{7}^{k} \int_{- \frac{3}{2}}^{- 1} z^{k - 4} dz

= - \sum_{k = 0 and k \neq 3}^{7} C_{7}^{k} {[\frac{1}{k - 3} z^{k - 3}]}_{- \frac{3}{2}}^{- 1} - C_{7}^{3} {[\ln ∣ z ∣]}_{- \frac{3}{2}}^{- 1}

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