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Question Number 98105 by mathmax by abdo last updated on 11/Jun/20

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

Answered by MJS last updated on 11/Jun/20

$$\int \frac{x+1}{{(x-2)}^2{(2x+3)}^3}dx=$$$$\left[\mathrm{Ostrogradski}\right]$$$$=-\frac{44x^2+55x+8}{686(x-2){(2x+3)}^2}-\frac{11}{343}\int \frac{dx}{(x-2)(2x+3)}=$$$$=-\frac{44x^2+55x+8}{686(x-2){(2x+3)}^2}+\frac{11}{2401}\ln \mid \frac{2x+3}{x-2}\mid +C$$$$\underset{3}{\stackrel{+\mathrm{\infty }}{\int }}\frac{x+1}{{(x-2)}^2{(2x+3)}^3}dx=\frac{569}{55566}+\frac{11}{2401}(\ln 2-2\ln 3)$$

Commented by mathmax by abdo last updated on 11/Jun/20

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

Answered by mathmax by abdo last updated on 11/Jun/20

$$\mathrm{binome}\mathrm{method}\mathrm{I}={\int }_3^{+\mathrm{\infty }}\frac{(\mathrm{x}+1)\mathrm{dx}}{{(\mathrm{x}-2)}^2{(2\mathrm{x}+3)}^3}\Rightarrow$$$$\mathrm{I}={\int }_3^{+\mathrm{\infty }}\frac{(\mathrm{x}+1)\mathrm{dx}}{{\left(\frac{\mathrm{x}-2}{2\mathrm{x}+3}\right)}^2{(2\mathrm{x}+3)}^5}\mathrm{changement}\frac{\mathrm{x}-2}{2\mathrm{x}+3}=\mathrm{t}\mathrm{give}\mathrm{x}-2=2\mathrm{tx}+3\mathrm{t}\Rightarrow$$$$(1-2\mathrm{t})\mathrm{x}=3\mathrm{t}+2\Rightarrow \mathrm{x}=\frac{3\mathrm{t}+2}{1-2\mathrm{t}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3(1-2\mathrm{t})-(3\mathrm{t}+2)(-2)}{{(1-2\mathrm{t})}^2}$$$$=\frac{3-6\mathrm{t}+6\mathrm{t}+4}{{(1-2\mathrm{t})}^2}=\frac{7}{{(1-2\mathrm{t})}^2},2\mathrm{x}+3=\frac{6\mathrm{t}+4}{1-2\mathrm{t}}+3=\frac{6\mathrm{t}+4+3-6\mathrm{t}}{1-2\mathbf{t}}=\frac{7}{1-2\mathbf{t}}$$$$\mathbf{x}+1=\frac{3\mathbf{t}+2}{1-2\mathbf{t}}+1=\frac{3\mathbf{t}+2+1-2\mathbf{t}}{1-2\mathbf{t}}=\frac{\mathbf{t}+3}{1-2\mathbf{t}}\Rightarrow$$$$\mathrm{I}={\int }_{\frac{1}{9}}^{\frac{1}{2}}\frac{(\mathrm{t}+3)}{(1-2\mathrm{t}){\mathrm{t}}^2{\left(\frac{7}{1-2\mathrm{t}}\right)}^5}\times \frac{7}{{(1-2\mathrm{t})}^2}\mathrm{dt}=\frac{1}{7^6}{\int }_{\frac{1}{9}}^{\frac{1}{2}}\frac{(\mathrm{t}+3){(1-2\mathrm{t})}^5}{{(1-2\mathrm{t})}^3{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{7^6}{\int }_{\frac{1}{9}}^{\frac{1}{2}}\frac{(\mathrm{t}+3){(1-2\mathrm{t})}^2}{{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{7^6}{\int }_{\frac{1}{9}}^{\frac{1}{2}}\frac{(\mathrm{t}+3)({4\mathrm{t}}^2-4\mathrm{t}+1)}{{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{7^6}{\int }_{\frac{1}{9}}^{\frac{1}{9}}\frac{{4\mathrm{t}}^3-{4\mathrm{t}}^2+\mathrm{t}+{12\mathrm{t}}^2-12\mathrm{t}+3}{{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{7^6}{\int }_{\frac{1}{9}}^{\frac{1}{9}}\frac{{4\mathrm{t}}^3+{8\mathrm{t}}^2-11\mathrm{t}+3}{{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{7^6}\int$$

Commented by mathmax by abdo last updated on 11/Jun/20

$$\mathrm{I}=\frac{1}{7^6}\left\{{\int }_{\frac{1}{9}}^{\frac{1}{2}}(4\mathrm{t}+8-\frac{11}{\mathrm{t}}+\frac{3}{{\mathrm{t}}^2})\mathrm{dt}\right\}$$$$=\frac{1}{7^6}{[{2\mathrm{t}}^2+8\mathrm{t}-11\ln \mid \mathrm{t}\mid -\frac{3}{\mathrm{t}}]}_{\frac{1}{9}}^{\frac{1}{2}}$$$$=\frac{1}{7^6}(\frac{1}{2}+4+11\ln \left(2\right)-6-2{\left(\frac{1}{9}\right)}^2-8\times \frac{1}{9}11\ln \left(\frac{1}{9}\right)+27)=...$$

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