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Question Number 65137 by Tony Lin last updated on 25/Jul/19

$$\int \frac{dx}{{(x-2)}^3{(x+1)}^2}=?$$

Commented by mathmax by abdo last updated on 25/Jul/19

$$letI=\int \frac{dx}{{(x-2)}^3{(x+1)}^2}changementx-2=tgive$$$$I=\int \frac{dt}{t^3{(t+3)}^2}letdecomposeF\left(t\right)=\frac{1}{t^3{(t+3)}^2}\Rightarrow$$$$F\left(t\right)=\frac{a}{t}+\frac{b}{t^2}+\frac{c}{t^3}+\frac{d}{t+3}+\frac{e}{{(t+3)}^2}$$$$c={lim}_{t\to 0}{t}^{3}F\left(t\right)=\frac{1}{9}$$$$e={lim}_{t\to -3}{(t+3)}^{2}F\left(t\right)=-\frac{1}{27}\Rightarrow F\left(t\right)=\frac{a}{t}+\frac{b}{t^2}+\frac{1}{9t^3}+\frac{d}{t+3}-\frac{1}{27{(t+3)}^2}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+d\Rightarrow d=-a\Rightarrow$$$$F\left(t\right)=\frac{a}{t}+\frac{b}{t^2}+\frac{1}{9t^3}-\frac{a}{t+3}-\frac{1}{27{(t+3)}^2}$$$$F\left(1\right)=\frac{1}{16}=a+b+\frac{1}{9}-\frac{a}{4}-\frac{1}{27.16}\Rightarrow 1=16a+16b+\frac{16}{9}-4a-\frac{1}{27}$$$$\Rightarrow 12a+16b+\frac{47}{27}=1\Rightarrow 12a+16b=1-\frac{47}{27}=-\frac{20}{27}\Rightarrow 3a+4b=-\frac{5}{27}$$$$F(-2)=-\frac{1}{8}=-\frac{a}{2}+\frac{b}{4}-\frac{1}{9.8}-a-\frac{1}{27}\Rightarrow$$$$\frac{1}{8}=\frac{a}{2}-\frac{b}{4}+\frac{1}{9.8}+a+\frac{1}{27}\Rightarrow 1=4a-2b+\frac{1}{9}+8a+\frac{8}{27}\Rightarrow$$$$12a-2b+\frac{11}{27}=1\Rightarrow 12a-2b=1-\frac{11}{27}=\frac{16}{27}\Rightarrow 6a-b=\frac{4}{27}\Rightarrow$$$$b=6a-\frac{4}{27}\Rightarrow 3a+4(6a-\frac{4}{27})=-\frac{5}{27}\Rightarrow 27a=\frac{16}{27}-\frac{5}{27}=\frac{11}{27}\Rightarrow a=\frac{11}{{27}^2}$$$$b=\frac{66}{{27}^2}-\frac{4.27}{{27}^2}=-\frac{42}{{27}^2}\Rightarrow F\left(t\right)=\frac{11}{{27}^{2}t}-\frac{42}{{27}^2{t}^2}+\frac{1}{9t^3}-\frac{11}{{27}^2(t+3)}-\frac{1}{27{(t+3)}^2}$$$$\Rightarrow I=\int F\left(t\right)dt=\frac{11}{{27}^2}ln\mid t\mid +\frac{42}{27t}-\frac{1}{18t^2}-\frac{11}{{27}^2}ln\mid t+3\mid +\frac{1}{27(t+3)}+C$$$$=\frac{11}{{27}^2}ln\mid x-2\mid +\frac{42}{27(x-2)}-\frac{1}{18{(x-2)}^2}-\frac{11}{{27}^2}ln\mid x+1\mid +\frac{1}{27(x+1)}+C.$$$$$$

Commented by Tony Lin last updated on 26/Jul/19

$$thankssir$$

Commented by mathmax by abdo last updated on 26/Jul/19

$$youarewelcomesir.$$

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