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

$$\mathrm{calculate}{\int }_2^{+\mathrm{\infty }}\frac{\mathrm{dt}}{{(2\mathrm{t}+3)}^4{(\mathrm{t}-1)}^5}$$

Answered by Olaf last updated on 17/Sep/20

$$\mathrm{R}\left(x\right)=$$$$\frac{\frac{1}{625}}{{(x-1)}^5}$$$$-\frac{\frac{8}{3125}}{{(x-1)}^4}$$$$+\frac{\frac{8}{3135}}{{(x-1)}^3}$$$$+\frac{\frac{32}{15625}}{{(x-1)}^2}$$$$+\frac{\frac{112}{78125}}{x-1}$$$$+\frac{\frac{32}{3125}}{{(2x+3)}^4}$$$$-\frac{\frac{32}{3125}}{{(2x+3)}^3}$$$$-\frac{\frac{96}{15625}}{{(2x+3)}^2}$$$$-\frac{\frac{225}{78125}}{2x+3}$$$${\int }_2^{\mathrm{\infty }}\mathrm{R}\left(x\right)dx=...$$

Answered by Olaf last updated on 17/Sep/20

$$u=t-1$$$$\mathrm{I}={\int }_1^{\mathrm{\infty }}\frac{du}{u^5{(u+5)}^4}$$$$x=\frac{1}{u}$$$$\mathrm{I}={\int }_1^0\frac{x^5}{{(\frac{1}{x}+5)}^5}(-\frac{dx}{x^2})$$$$\mathrm{I}={\int }_0^1\frac{x^8}{{(5x+1)}^5}dx$$$$\lambda =5x+1$$$$\mathrm{I}=\frac{1}{5^8}{\int }_1^6\frac{{(\lambda -1)}^8}{{\lambda }^5}.\frac{d\lambda }{5}$$$$\mathrm{I}=\frac{1}{5^9}{\int }_1^6\frac{\underset{k=0}{\sum ^8}{\mathrm{C}}_8^k{\lambda }^k{(-1)}^{8-k}}{{\lambda }^5}d\lambda$$$$...maybethiswayisbetter...$$$$$$

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

$$\mathrm{Put}\mathrm{u}=\frac{1}{\mathrm{t}-1}\Rightarrow \mathrm{du}=\frac{-1}{{(\mathrm{t}-1)}^2}\mathrm{dt}$$$$\mathrm{dt}=-{(\mathrm{t}-1)}^2\mathrm{du}=\frac{-1}{{\mathrm{u}}^2}\mathrm{du},\mathrm{t}=\frac{1}{\mathrm{u}}+1$$$${\int }_2^{\mathrm{\infty }}\frac{\mathrm{dt}}{{(2\mathrm{t}+3)}^4{(\mathrm{t}-1)}^5}={\int }_1^0\frac{-{\mathrm{u}}^3\mathrm{du}}{{(\frac{2}{\mathrm{u}}+5)}^4}$$$$=-{\int }_1^0\frac{{\mathrm{u}}^7}{{(5\mathrm{u}+2)}^4}\mathrm{du}=-\frac{1}{625}{\int }_{7/5}^0\frac{{\mathrm{u}}^7}{{(\mathrm{u}+\frac{2}{5})}^4}\mathrm{du}$$$$\mathrm{Put}\mathrm{u}+\frac{2}{5}=\mathrm{v}\Rightarrow \mathrm{du}=\mathrm{dv},\mathrm{u}=\mathrm{v}-\frac{2}{5}$$$$\mathrm{I}=-\frac{1}{625}{\int }_{7/5}^{2/5}\frac{{(\mathrm{v}-\frac{2}{5})}^7}{{\mathrm{v}}^4}\mathrm{dv}$$$$=\frac{-1}{625}{\int }_{7/5}^{2/5}\{[{\mathrm{v}}^7-{7\mathrm{v}}^6.0.4+{21\mathrm{v}}^5.0.4^2$$$$-{35\mathrm{v}}^4.0.4^3+{35\mathrm{v}}^3.0.4^4-{21\mathrm{v}}^2.{(0.4)}^5$$$$+7\mathrm{v}.{(0.4)}^6-{(0.4)}^7]/{\mathrm{v}}^4\}\mathrm{dv}$$$$=\frac{-1}{625}{\int }_{7/5}^{\mathrm{\infty }}[{\mathrm{v}}^3-\frac{14}{5}{\mathrm{v}}^2+21\mathrm{v}.{\left(\frac{2}{5}\right)}^2-\frac{35.8}{125}$$$$+\frac{35.16}{625\mathrm{v}}-\frac{21.32}{{3125\mathrm{v}}^2}+\frac{7.64}{{15625\mathrm{v}}^3}-\frac{128}{5^7{\mathrm{v}}^4}]$$$$=\frac{-1}{625}\{\frac{{\mathrm{v}}^4}{4}-\frac{{14\mathrm{v}}^3}{15}+\frac{42}{25}{\mathrm{v}}^2-\frac{280}{125}\mathrm{v}$$$${+\frac{560}{625}.\ln \mid \mathrm{v}\mid +\frac{672}{3125\mathrm{v}}-\frac{224}{{15625\mathrm{v}}^2}+\frac{128}{3.5^7{\mathrm{v}}^3}]}_{7/5}^{2/5}$$$$=-\frac{1}{625}[(-1.0449965-(-0.99590280)]=$$$$(-0.0490937)/625=0.00007855$$$$$$

Answered by MJS_new last updated on 17/Sep/20

$$\int \frac{dt}{{(2t+3)}^4{(t-1)}^5}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{5376t^6+1344t^5-25760t^4+3080t^3+40740t^2-17906t-16249}{187500{(t-1)}^4{(2t+3)}^3}+$$$$+\frac{112}{15625}\int \frac{dt}{(t-1)(2t+3)}$$$$\mathrm{this}\mathrm{integral}=\frac{112}{78125}\ln \mid \frac{t-1}{2t+3}\mid$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{112}{78125}(\ln 7-\ln 2)-\frac{36817}{21437500}$$

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

$$\mathrm{Great}!{\mathrm{Sir}}^{\prime }\mathrm{s}\mathrm{result}\mathrm{coincde}\mathrm{to}\mathrm{my}\mathrm{result}$$

Commented by MJS_new last updated on 17/Sep/20

$$\mathrm{yes}.$$$$\mathrm{I}\mathrm{like}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\mathrm{better}\mathrm{than}$$$$\mathrm{decomposing}...$$

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