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Question Number 56939 by maxmathsup by imad last updated on 26/Mar/19

$$calculate\int \frac{dx}{{(x+1)}^3(x^2-3x+2)}$$$$2)findthevalueof{\int }_2^{+\mathrm{\infty }}\frac{dx}{{(x+1)}^3(x^2-3x+2)}$$

Commented by turbo msup by abdo last updated on 27/Mar/19

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

Commented by maxmathsup by imad last updated on 27/Mar/19

$$changementx+1=tgiveI=\int \frac{dt}{t^3({(t-1)}^2-3(t-1)+2)}$$$$=\int \frac{dt}{t^3(t^2-2t+1-3t+3+2)}=\int \frac{dt}{t^3(t^2-5t+6)}letdecopose$$$$F\left(t\right)=\frac{1}{t^3(t^2-5t+6)}rootsof{t}^2-5t+6$$$$\mathrm{\Delta }=25-24=1\Rightarrow t_1=\frac{5+1}{2}=3and{t}_2=\frac{5-1}{2}=2\Rightarrow$$$$F\left(t\right)=\frac{1}{t^3(t-3)(t-2)}=\frac{a}{t}+\frac{b}{t^2}+\frac{c}{t^3}+\frac{d}{t-3}+\frac{e}{t-2}$$$$c={lim}_{t\to 0}{t}^{3}F\left(t\right)=\frac{1}{6}$$$$d={lim}_{t\to 3}(t-3)F\left(t\right)=\frac{1}{27}$$$$e={lim}_{t\to 2}(t-2)F\left(t\right)=-\frac{1}{8}\Rightarrow F\left(t\right)=\frac{a}{t}+\frac{b}{t^2}+\frac{1}{6t^3}+\frac{1}{27(t-3)}-\frac{1}{8(t-2)}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+\frac{1}{27}-\frac{1}{8}=a+\frac{8-27}{27.8}=a-\frac{19}{216}\Rightarrow a=\frac{19}{216}\Rightarrow$$$$F\left(t\right)=\frac{19}{216t}+\frac{b}{t^2}+\frac{1}{6t^3}+\frac{1}{27(t-3)}-\frac{1}{8(t-2)}$$$$F\left(1\right)=\frac{1}{2}=\frac{19}{216}+b+\frac{1}{6}-\frac{1}{54}+\frac{1}{8}\Rightarrow 1=\frac{19}{108}+2b+\frac{1}{3}-\frac{1}{27}+\frac{1}{4}$$$$=2b+\frac{19}{108}+\frac{7}{12}-\frac{1}{27}\Rightarrow 2b=1-\frac{19}{108}-\frac{7}{12}+\frac{1}{27}\Rightarrow b=\frac{1}{2}-\frac{19}{216}-\frac{7}{24}+\frac{1}{54}=b_0\Rightarrow$$$$\int F\left(t\right)dt=\frac{19}{216}ln\mid t\mid -\frac{b_0}{t}+\frac{1}{6}\frac{1}{-2}{t}^{-2}+\frac{1}{27}ln\mid t-3\mid -\frac{1}{8}ln\mid t-2\mid +c$$$$=\frac{19}{216}ln\mid t\mid -\frac{b_0}{t}-\frac{1}{12t^2}+\frac{1}{27}ln\mid t-3\mid -\frac{1}{8}ln\mid t-2\mid +c$$$$=\frac{19}{216}ln\mid x+1\mid -\frac{b_0}{x+1}-\frac{1}{12{(x+1)}^2}+\frac{1}{27}ln\mid x-2\mid -\frac{1}{8}ln\mid x-1\mid +c=I.$$$$$$

Answered by MJS last updated on 27/Mar/19

$$\int \frac{dx}{{(x+1)}^3(x^2-3x+2)}=\int \frac{dx}{(x-2)(x-1){(x+1)}^3}=$$$$=\int (\frac{1}{27(x-2)}-\frac{1}{8(x-1)}+\frac{19}{216(x+1)}+\frac{5}{36{(x+1)}^2}+\frac{1}{6{(x+1)}^3})dx=$$$$=\frac{1}{27}\ln \mid x-2\mid -\frac{1}{8}\ln \mid x-1\mid +\frac{19}{216}\ln \mid x+1\mid -\frac{5}{36(x+1)}-\frac{1}{12{(x+1)}^2}+C=$$$$=\frac{1}{27}\ln \mid x-2\mid -\frac{1}{8}\ln \mid x-1\mid +\frac{19}{216}\ln \mid x+1\mid -\frac{5x+8}{36{(x+1)}^2}+C$$$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{integral}\mathrm{in}[2;+\mathrm{\infty }[{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented by turbo msup by abdo last updated on 27/Mar/19

$$thankyousir.$$

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