calculate ∫_0 ^∞ ((x^4 dx)/((2x+1)^5 (3x+1)^8 ))


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Question Number 119762 by mathmax by abdo last updated on 26/Oct/20

$calculate \int_{0}^{\infty} \frac{x^{4} dx}{{(2 x + 1)}^{5} {(3 x + 1)}^{8}}$
	Answered by Olaf last updated on 26/Oct/20

	$\frac{5319686}{105} + 124952 \ln (\frac{2}{3})$ $. . . The calculous is very long!$ $R (x) = \frac{x^{4}}{{(2 x + 1)}^{5} {(3 x + 1)}^{8}}$ $R (x) = \frac{16}{{(2 x + 1)}^{5}} + \frac{320}{{(2 x + 1)}^{4}} + \frac{3744}{{(2 x + 1)}^{3}}$ $+ \frac{37344}{{(2 x + 1)}^{2}} + \frac{249904}{2 x + 1} + \frac{3}{{(3 x + 1)}^{8}}$ $- \frac{42}{{(3 x + 1)}^{7}} + \frac{318}{{(3 x + 1)}^{6}} - \frac{1752}{{(3 x + 1)}^{5}}$ $+ \frac{7923}{{(3 x + 1)}^{4}} - \frac{31326}{{(3 x + 1)}^{3}} + \frac{112404}{{(3 x + 1)}^{2}}$ $- \frac{374856}{3 x + 1}$ $. . .$
	Commented by mathmax by abdo last updated on 26/Oct/20

	$thanks sir$
	Answered by MJS_new last updated on 26/Oct/20

	$method (1) decomposition$ $method (2) Ostrogradski$ $both are long . . .$ $method (3)$ $\int \frac{x^{4}}{{(2 x + 1)}^{5} {(3 x + 1)}^{8}} d x =$ $[t = \frac{2 x + 1}{3 x + 1} \to d x = - {(3 x + 1)}^{2} d t]$ $= - \int \frac{{(t - 1)}^{4} {(3 t - 2)}^{7}}{t^{5}} d t$ $this seems easier to me . . .$ $\frac{{(t - 1)}^{4} {(3 t - 2)}^{7}}{t^{5}} =$ $= 2187 t^{6} - 18954 t^{5} + 74358 t^{4} - 174312 t^{3} +$ $+ 271323 t^{2} - 294462 t - \frac{124952}{t} + \frac{47888}{t^{2}} -$ $- \frac{12192}{t^{3}} + \frac{1856}{t^{4}} - \frac{128}{t^{5}} + 227388$ $and these are easy to integrate$
	Commented by mathmax by abdo last updated on 27/Oct/20

	$thank you sir mjs$
	Commented by MJS_new last updated on 27/Oct/20

	$as always, you are welcome$
	Answered by mathmax by abdo last updated on 27/Oct/20

	$let I = \int_{0}^{\infty} \frac{x^{4}}{{(2 x + 1)}^{5} {(3 x + 1)}^{8}} dx \Rightarrow$ $I = \int_{0}^{\infty} \frac{x^{4}}{{(\frac{2 x + 1}{3 x + 1})}^{5} {(3 x + 1)}^{13}} dx we do the chang . \frac{2 x + 1}{3 x + 1} = t \Rightarrow$ $2 x + 1 = 3 tx + t \Rightarrow (2 - 3 t) x = t - 1 \Rightarrow x = \frac{t - 1}{2 - 3 t} \Rightarrow$ $\frac{dx}{dt} = \frac{2 - 3 t - (t - 1) (- 3)}{{(2 - 3 t)}^{2}} = \frac{2 - 3 t + 3 t - 3}{{(2 - 3 t)}^{2}} = \frac{- 1}{{(2 - 3 t)}^{2}}$ $3 x + 1 = \frac{3 t - 3}{2 - 3 t} + 1 = \frac{3 t - 3 + 2 - 3 t}{2 - 3 t} = \frac{- 1}{2 - 3 t} \Rightarrow$ $I = \int_{1}^{\frac{2}{3}} \frac{{(t - 1)}^{4}}{{(2 - 3 t)}^{4}} \times \frac{1}{t^{5} {(\frac{- 1}{2 - 3 t})}^{13}} \times \frac{- 1}{{(2 - 3 t)}^{2}} dt$ $= \int_{\frac{2}{3}}^{1} \frac{{(t - 1)}^{4}}{{(2 - 3 t)}^{6} t^{5}} \times {(3 t - 2)}^{13} dt$ $= \int_{\frac{2}{3}}^{1} \frac{{(t - 1)}^{4} {(3 t - 2)}^{7}}{t^{5}} dt we have$ ${(t - 1)}^{4} = \sum_{k = 0}^{4} C_{4}^{k} t^{k} {(- 1)}^{k} = 1 - C_{4}^{1} t + C_{4}^{2} t^{2} - C_{4}^{3} t^{3} + t^{4}$ $= 1 - 4 t + {6 t}^{2} - {4 t}^{3} + t^{4} \Rightarrow$ $I = \int_{\frac{2}{3}}^{1} (\frac{1}{t^{5}} - \frac{4}{t^{4}} + \frac{6}{t^{3}} - \frac{4}{t^{2}} + \frac{1}{t}) {(3 t - 2)}^{7} dt$ $= \int_{\frac{2}{3}}^{1} \frac{{(3 t - 2)}^{7}}{t^{5}} dt - 4 \int_{\frac{2}{3}}^{1} \frac{{(3 t - 2)}^{7}}{t^{4}} + 6 \int_{\frac{2}{3}}^{1} \frac{{(3 t - 2)}^{7}}{t^{3}} dt - 4 \int_{\frac{2}{3}}^{1} \frac{{(3 t - 2)}^{7}}{t^{2}}$ $+ \int_{\frac{2}{3}}^{1} \frac{{(3 t - 2)}^{7}}{t} dt those integrals can be calculsted by binome formula$ $\int_{\frac{2}{3}}^{1} \frac{{(3 t - 2)}^{7}}{t^{5}} dt = \int_{\frac{2}{3}}^{1} \frac{\sum_{k = 0}^{7} C_{7}^{k} {(3 t)}^{k} {(- 2)}^{7 - k}}{t^{5}} dt$ $= \int_{\frac{2}{3}}^{1} \sum_{k = 0}^{7} C_{7}^{k} 3^{k} {(- 2)}^{7 - k} t^{k - 5} dt$ $= \sum_{k = 0 and k \neq 4}^{7} C_{7}^{k} 3^{k} {(- 2)}^{7 - k} {[\frac{1}{k - 4} t^{k - 4}]}_{\frac{2}{3}}^{1} + C_{7}^{4} 3^{4} {(- 2)}^{3} {[\ln ∣ t ∣]}_{\frac{2}{3}}^{1}$ $= \sum_{k = 0 and k \neq 4}^{7} \frac{C_{7}^{k} 3^{k} {(- 2)}^{7 - k}}{k - 4} (1 - {(\frac{2}{3})}^{k - 4}) - C_{7}^{4} 3^{4} {(- 2)}^{3} \ln (\frac{2}{3})$ $. . . .$
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