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Question Number 103591 by mathmax by abdo last updated on 16/Jul/20

$$\mathrm{calculate}{\int }_3^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2-1)}^3{(\mathrm{x}+2)}^2}$$

Commented by Worm_Tail last updated on 16/Jul/20

$$$$$${\int }_3^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2-1)}^3{(\mathrm{x}+2)}^2}pfd$$$${\int }_3^{oo}\frac{-3}{16(x+1)}+\frac{4}{27(x+2)}+\frac{1}{16{(x+1)}^2}+\frac{1}{27{(x+2)}^2}-\frac{1}{8{(x+1)}^3}+\frac{17}{432(x-1)}-\frac{13}{432{(x-1)}^2}+\frac{1}{72{(x-1)}^3}dx$$$${[\frac{-3ln(x+1)}{16}+\frac{4}{27}ln(x+2)-\frac{1}{16(x+1)}-\frac{1}{27(x+2)}+\frac{1}{16{(x+1)}^2}+\frac{17ln(x-1)}{432}+\frac{13}{432(x-1)}-\frac{1}{144{(x-1)}^2}]}_3^{oo}$$$${[\frac{ln\frac{{(x+2)}^{64}{(x-1)}^{17}}{{(x+1)}^{81}}}{432}-\frac{1}{27(x+2)}+\frac{1}{16{(x+1)}^2}+\frac{13}{432(x-1)}-\frac{1}{144{(x-1)}^2}]}_3^{oo}$$$$li\underset{t\to oo}{m}[\frac{ln\frac{{(t+2)}^{64}{(t-1)}^{17}}{{(t+1)}^{81}}}{432}-\frac{1}{27(t+2)}+\frac{1}{16{(t+1)}^2}+\frac{13}{432(t-1)}-\frac{1}{144{(t-1)}^2}]$$$$-[\frac{ln\frac{{(3+2)}^{64}{(3-1)}^{17}}{{(3+1)}^{81}}}{432}-\frac{1}{27(3+2)}+\frac{1}{16{(3+1)}^2}+\frac{13}{432(3-1)}-\frac{1}{144{(3-1)}^2}]$$$$$$$$li\underset{t\to oo}{m}[\frac{ln\left(1\right)}{432}-0+0+0-0]-[\frac{ln\frac{5^{64}\times 2^{17}}{4^{81}}}{432}+\frac{113}{11520}]$$$$-[\frac{64ln\left(5\right)-145ln\left(2\right)}{432}+\frac{113}{11520}]$$$$\frac{-64ln\left(5\right)+145ln\left(2\right)}{432}-\frac{113}{11520}$$$$$$$$$$

Commented by Worm_Tail last updated on 16/Jul/20

$$mistakemighthavecreptin$$

Commented by abdomathmax last updated on 16/Jul/20

$$\mathrm{can}\mathrm{you}\mathrm{chow}\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{find}\mathrm{the}\mathrm{decomposition}?$$

Commented by Worm_Tail last updated on 16/Jul/20

$$$$$${\int }_3^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2-1)}^3{(\mathrm{x}+2)}^2}pfd$$$$\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{{(x+1)}^2}+\frac{D}{{(x+2)}^2}+\frac{E}{{(x+1)}^3}+\frac{F}{(x-1)}+\frac{G}{{(x-1)}^2}+\frac{H}{{(x-1)}^3}=\frac{1}{{(x^2-1)}^3{(x+2)}^2}$$$$youcontinuefromhere$$$$$$$$$$$$$$

Answered by abdomathmax last updated on 16/Jul/20

$$\mathrm{I}={\int }_3^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2-1)}^3{(\mathrm{x}+2)}^2}\Rightarrow \mathrm{I}={\int }_3^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}-1)}^3{(\mathrm{x}+1)}^3{(\mathrm{x}+2)}^2}$$$$={\int }_3^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^3{(\mathrm{x}+1)}^6{(\mathrm{x}+2)}^2}\mathrm{we}\mathrm{do}\mathrm{tbe}\mathrm{cha}7\mathrm{gement}$$$$\frac{\mathrm{x}-1}{\mathrm{x}+1}=\mathrm{t}\Rightarrow \mathrm{x}-1=\mathrm{tx}+\mathrm{t}\Rightarrow (1-\mathrm{t})\mathrm{x}=\mathrm{t}+1\Rightarrow$$$$\mathrm{x}=\frac{1+\mathrm{t}}{1-\mathrm{t}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1-\mathrm{t}+(1+\mathrm{t})}{{(1-\mathrm{t})}^2}=\frac{2}{{(1-\mathrm{t})}^2}$$$$\mathrm{x}+1=\frac{1+\mathrm{t}}{1-\mathrm{t}}+1=\frac{1+\mathrm{t}+1-\mathrm{t}}{1-\mathrm{t}}=\frac{2}{1-\mathrm{t}}$$$$\mathrm{x}+2=\frac{1+\mathrm{t}}{1-\mathrm{t}}+2=\frac{1+\mathrm{t}+2-2\mathrm{t}}{1-\mathrm{t}}=\frac{3-\mathrm{t}}{1-\mathrm{t}}\Rightarrow$$$$\mathrm{I}={\int }_{\frac{1}{2}}^1\frac{2\mathrm{dt}}{{(1-\mathrm{t})}^2{\mathrm{t}}^3{\left(\frac{2}{1-\mathrm{t}}\right)}^6{\left(\frac{3-\mathrm{t}}{1-\mathrm{t}}\right)}^2}$$$$=2{\int }_{\frac{1}{2}}^1\frac{{(\mathrm{t}-1)}^8}{{(\mathrm{t}-1)}^2{\mathrm{t}}^3.2^6{(3-\mathrm{t})}^2}\mathrm{dt}$$$$=\frac{1}{2^5}{\int }_{\frac{1}{2}}^1\frac{{(\mathrm{t}-1)}^6}{{\mathrm{t}}^3{(3-\mathrm{t})}^2}\mathrm{dt}=\frac{1}{2^5}{\int }_{\frac{1}{2}}^1\frac{{(\mathrm{t}-1)}^6}{{\left(\frac{\mathrm{t}}{\mathrm{t}-3}\right)}^3{(\mathrm{t}-3)}^5}\mathrm{dt}$$$$\mathrm{again}\mathrm{ch}.\frac{\mathrm{t}}{\mathrm{t}-3}=\mathrm{z}\Rightarrow \mathrm{t}=\mathrm{zt}-3\mathrm{z}\Rightarrow (1-\mathrm{z})\mathrm{t}=-3\mathrm{z}\Rightarrow$$$$\mathrm{t}=\frac{-3\mathrm{z}}{1-\mathrm{z}}=\frac{3\mathrm{z}}{\mathrm{z}-1}\mathrm{and}\mathrm{t}-3=\frac{3\mathrm{z}}{\mathrm{z}-1}-3=\frac{3\mathrm{z}-3\mathrm{z}+3}{\mathrm{z}-1}$$$$=\frac{3}{\mathrm{z}-1}\mathrm{and}\mathrm{t}-1=\frac{3\mathrm{z}}{\mathrm{z}-1}-1=\frac{3\mathrm{z}-\mathrm{z}+1}{\mathrm{z}-1}=\frac{2\mathrm{z}+1}{\mathrm{z}-1}$$$$\mathrm{I}=\frac{1}{2^5}{\int }_{-\frac{1}{5}}^{-\frac{1}{2}}\frac{{\left(\frac{2\mathrm{z}+1}{\mathrm{z}-1}\right)}^6}{{\mathrm{z}}^3{\left(\frac{3}{\mathrm{z}-1}\right)}^5}\times \frac{-3}{{(\mathrm{z}-1)}^2}\mathrm{dz}$$$$=-\frac{1}{2^5.3^4}{\int }_{-\frac{1}{5}}^{-\frac{1}{2}}\frac{{(2\mathrm{z}+1)}^6{(\mathrm{z}-1)}^5}{{\mathrm{z}}^3{(\mathrm{z}-1)}^8}\mathrm{dz}$$$$=-\frac{1}{2^5.3^4}{\int }_{-\frac{1}{5}}^{-\frac{1}{2}}\frac{{(2\mathrm{z}+1)}^6}{{\mathrm{z}}^3{(\mathrm{z}-1)}^3}\mathrm{dz}$$$$=-\frac{1}{2^5.3^4}{\int }_{-\frac{1}{5}}^{-\frac{1}{2}}\frac{\sum _{\mathrm{k}=0}^6{\mathrm{C}}_6^{\mathrm{k}}{\left(2\mathrm{z}\right)}^{\mathrm{k}}}{{\mathrm{z}}^3{(\mathrm{z}-1)}^3}\mathrm{dz}$$$$=-\frac{1}{2^5.3^4}{\int }_{-\frac{1}{5}}^{-\frac{1}{2}}\sum _{\mathrm{k}=0}^6{2}^{\mathrm{k}}{\mathrm{C}}_6^{\mathrm{k}}\frac{{\mathrm{z}}^{\mathrm{k}}}{{\mathrm{z}}^3{(\mathrm{z}-1)}^3}\mathrm{dz}$$$$=-\frac{1}{2^5.3^4}\sum _{\mathrm{k}=0}^6{2}^{\mathrm{k}}{\mathrm{C}}_6^{\mathrm{k}}{\int }_{-\frac{1}{5}}^{-\frac{1}{2}}\frac{{\mathrm{z}}^{\mathrm{k}}}{{\mathrm{z}}^3{(\mathrm{z}-1)}^3}\mathrm{dz}$$$$\mathrm{after}\mathrm{we}\mathrm{decompose}{\mathrm{F}}_{\mathrm{k}}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^{\mathrm{k}}}{{\mathrm{z}}^3{(\mathrm{z}-1)}^3}$$$$...\mathrm{be}\mathrm{continued}...$$$$$$$$$$$$$$

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