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Question Number 137694 by Mathspace last updated on 05/Apr/21

$$find\int \frac{x^3}{{(x-3)}^3{(x\sqrt{2}+7)}^4}dx$$

Commented by MJS_new last updated on 05/Apr/21

$$\mathrm{you}\mathrm{can}\mathrm{either}\mathrm{decompose}\mathrm{or}\mathrm{apply}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}$$$$\mathrm{Method};{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{fun}\mathrm{typing}\mathrm{with}\mathrm{these}\mathrm{constants}...$$

Commented by MJS_new last updated on 05/Apr/21

$$\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{to}\mathrm{study}\mathrm{the}\mathrm{methods}\mathrm{try}$$$$\int \frac{x^3}{{(x-1)}^3{(x+1)}^4}dx=$$$$=-\frac{(x^2+x+4)(3x^2-1)}{48{(x-1)}^2{(x+1)}^3}+\frac{1}{32}\ln \mid \frac{x+1}{x-1}\mid +C$$$$\mathrm{your}\mathrm{example}\mathrm{uses}\mathrm{the}\mathrm{same}\mathrm{path}\mathrm{with}\mathrm{nasty}\mathrm{constants}$$$$\int \frac{x^3}{{(x-3)}^3{(\sqrt{2}x+7)}^4}dx=$$$$=\frac{(1109519964-784602612\sqrt{2})x^4-(18723385548-13239011439\sqrt{2})x^3+(116325008541-82282349527\sqrt{2})x^2-(314307560196-222235080396\sqrt{2})x+311134260024-219840177123\sqrt{2}}{171774906{(x-3)}^2{(\sqrt{2}x+7)}^3}+\frac{9(115502365-81673862\sqrt{2})}{887503681}\ln \mid \frac{x-3}{\sqrt{2}x+7}\mid +C$$$$\mathrm{please}\mathrm{check}\mathrm{for}\mathrm{typos}\mathrm{if}\mathrm{you}\mathrm{can}$$

Commented by mathmax by abdo last updated on 06/Apr/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 06/Apr/21

$$\mathrm{I}=\int \frac{{\mathrm{x}}^3}{{(\mathrm{x}-3)}^3{(\mathrm{x}\sqrt{2}+7)}^4}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{{\mathrm{x}}^3}{{\left(\frac{\mathrm{x}-3}{\mathrm{x}\sqrt{2}+7}\right)}^3{(\mathrm{x}\sqrt{2}+7)}^7}\mathrm{dx}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{\mathrm{x}-3}{\mathrm{x}\sqrt{2}+7}=\mathrm{t}\Rightarrow \mathrm{x}-3=\sqrt{2}\mathrm{tx}+7\mathrm{t}\Rightarrow$$$$(1-\sqrt{2}\mathrm{t})\mathrm{x}=7\mathrm{t}+3\Rightarrow \mathrm{x}=\frac{7\mathrm{t}+3}{1-\sqrt{2}\mathrm{t}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{7(1-\sqrt{2}\mathrm{t})-(7\mathrm{t}+3)(-\sqrt{2})}{{(1-\sqrt{2}\mathrm{t})}^2}$$$$=\frac{7+3\sqrt{2}}{{(1-\sqrt{2}\mathrm{t})}^2}\mathrm{also}\mathrm{x}\sqrt{2}+7=\sqrt{2}.\frac{7\mathrm{t}+3}{1-\sqrt{2}\mathrm{t}}+7=\frac{7\sqrt{2}\mathrm{t}+3\sqrt{2}+7-7\sqrt{2}\mathrm{t}}{1-\sqrt{2}\mathrm{t}}$$$$=\frac{3\sqrt{2}+7}{1-\sqrt{2}\mathrm{t}}\Rightarrow \mathrm{I}=\int \frac{1}{{\mathrm{t}}^3{\left(\frac{3\sqrt{2}+7}{1-\sqrt{2}\mathrm{t}}\right)}^7}.\frac{3\sqrt{2}+7}{{(1-\sqrt{2}\mathrm{t})}^2}{\left(\frac{7\mathrm{t}+3}{1-\sqrt{2}\mathrm{t}}\right)}^3\mathrm{dt}$$$$=\frac{1}{{(3\sqrt{2}+7)}^6}\int \frac{{(1-\sqrt{2}\mathrm{t})}^7.{(7\mathrm{t}+3)}^3}{{(1-\sqrt{2}\mathrm{t})}^3.{\mathrm{t}}^3}\mathrm{dt}$$$$=\frac{1}{{(3\sqrt{2}+7)}^6}\int \frac{{(1-\sqrt{2}\mathrm{t})}^4{(7\mathrm{t}+3)}^3}{{\mathrm{t}}^3}\mathrm{dt}\mathrm{we}\mathrm{have}$$$${(1-\sqrt{2}\mathrm{t})}^4={(1-2\sqrt{2}\mathrm{t}+{2\mathrm{t}}^2)}^2={({2\mathrm{t}}^2-2\sqrt{2}\mathrm{t}+1)}^2$$$$={4\mathrm{t}}^4+{8\mathrm{t}}^2+1-4\sqrt{2}{\mathrm{t}}^3+{2\mathrm{t}}^2-2\sqrt{2}\mathrm{t}\mathrm{and}$$$${(7\mathrm{t}+3)}^3=7^3{\mathrm{t}}^3+3.{\left(7\mathrm{t}\right)}^2.3+3\left(7\mathrm{t}\right).3^2+3^3$$$$=343{\mathrm{t}}^3+{441\mathrm{t}}^2+189\mathrm{t}+27$$$$\Rightarrow {(1-\sqrt{2}\mathrm{t})}^4=({4\mathrm{t}}^4-4\sqrt{2}{\mathrm{t}}^3+{10\mathrm{t}}^2-2\sqrt{2}\mathrm{t}+1)({343\mathrm{t}}^3+{441\mathrm{t}}^2+189\mathrm{t}+27)$$$$..\mathrm{rest}\mathrm{to}\mathrm{finish}\mathrm{the}\mathrm{calculus}...$$$$$$

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