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Question Number 81313 by M±th+et£s last updated on 11/Feb/20

$$\int \frac{{(\mathrm{x}-2)}^3}{{(\mathrm{x}+2)}^5}\mathrm{dx}$$

Commented by M±th+et£s last updated on 11/Feb/20

$$\mathrm{sorry}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{typo}\mathrm{its}{(\mathrm{x}-2)}^{-3}$$

Commented by MJS last updated on 11/Feb/20

$$\int \frac{dx}{{(x+2)}^5{(x-2)}^3}=$$$$\mathrm{just}\mathrm{decompose}\mathrm{and}\mathrm{solve}.\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{want}\mathrm{to}$$$$\mathrm{do}\mathrm{this}\mathrm{job}\mathrm{for}\mathrm{you}$$$$=\int (\frac{A}{{(x+2)}^5}+\frac{B}{{(x+2)}^4}+\frac{C}{{(x+2)}^3}+\frac{D}{{(x+2)}^2}+\frac{E}{x+2}+\frac{F}{{(x-2)}^3}+\frac{G}{{(x-2)}^2}+\frac{H}{x-2})dx$$

Commented by Tony Lin last updated on 12/Feb/20

$$youcanletu=\frac{x+2}{x-2},\frac{du}{dx}=\frac{-4}{{(x-2)}^2}$$$$x=\frac{2u+2}{u-1}$$$$\Rightarrow -\frac{1}{4}\int \frac{du}{{\left(\frac{4u}{u-1}\right)}^5\left(\frac{4}{u-1}\right)}$$$$=-\frac{1}{4096}\int \frac{{(u-1)}^6}{u^5}du$$$$=-\frac{1}{4096}\int (u+\frac{15}{u}-\frac{20}{u^2}+\frac{15}{u^3}-\frac{6}{u^4}+\frac{1}{u^5}-6)du$$$$=-\frac{1}{4096}(\frac{u^2}{2}+15ln\mid u\mid +\frac{20}{u}-\frac{15}{2u^2}+\frac{2}{u^3}$$$$-\frac{1}{4u^4}-6u)+c$$$$plugu=\frac{x+2}{x-2}in$$$$\Rightarrow \int \frac{dx}{{(x+2)}^5{(x-2)}^3}$$$$=-\frac{1}{4096}[\frac{{\left(\frac{x+2}{x-2}\right)}^2}{2}+15ln\mid \frac{x+2}{x-2}\mid +\frac{20(x-2)}{x+2}$$$$-\frac{15}{2}{\left(\frac{x-2}{x+2}\right)}^2+2{\left(\frac{x-2}{x+2}\right)}^3-\frac{1}{4}{\left(\frac{x-2}{x+2}\right)}^4-$$$$\frac{6(x+2)}{x-2}]+c$$

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

$$I=\int \frac{dx}{{(x-2)}^3{(x+2)}^5}=\int \frac{dx}{{\left(\frac{x-2}{x+2}\right)}^3{(x+2)}^8}changement\frac{x-2}{x+2}=t$$$$givex-2=xt+2t\Rightarrow (1-t)x=2t+2\Rightarrow x=\frac{2t+2}{1-t}\Rightarrow x+2=2+\frac{2t+2}{1-t}$$$$=\frac{2-2t+2t+2}{1-t}=\frac{4}{(1-t)}$$$$dx=\frac{2(1-t)-(2t+2)(-1)}{{(1-t)}^2}dt=\frac{2-2t+2t+2}{{(t-1)}^2}=\frac{4}{{(t-1)}^2}dt\Rightarrow$$$$I=\int \frac{4dt}{{(t-1)}^2.t^3{\left(\frac{4}{1-t}\right)}^8}=\frac{1}{4^7}\int \frac{{(t-1)}^8}{{(t-1)}^2.t^3}dt$$$$=\frac{1}{4^7}\int \frac{{(t-1)}^4}{t^3}dt=\frac{1}{4^7}\int \frac{\sum _{k=0}^4{C}_4^k{t}^k{(-1)}^{4-k}}{t^3}dt$$$$=\frac{1}{4^7}\int \frac{C_4^0{(-1)}^4-C_4^{1}t+C_4^2{t}^2-C_4^3{t}^3+C_4^4{t}^4}{t^3}dt$$$$=\frac{1}{4^7}\int \frac{dt}{t^3}-\frac{C_4^1}{4^7}\int \frac{dt}{t^2}+\frac{C_4^2}{4^7}\int \frac{dt}{t}-\frac{C_4^3}{4^7}\int dt+\frac{1}{4^7}\int tdt$$$$=-\frac{1}{2}\frac{1}{4^7}\times \frac{1}{t^2}+\frac{1}{4^{6}t}+\frac{C_4^2}{4^7}ln\mid t\mid -\frac{1}{4^6}t+\frac{1}{2.4^7}{t}^2+C$$$$C_4^2=\frac{4!}{2!\times 2!}=\frac{4\times 3\times 2!}{{(2!)}^2}=6\Rightarrow$$$$I=-\frac{1}{2\times 4^7}{\left(\frac{x+2}{x-2}\right)}^2+\frac{1}{4^6}\left(\frac{x+2}{x-2)}\right)+\frac{6}{4^7}ln\mid \frac{x-2}{x+2}\mid -\frac{1}{4^6}\left(\frac{x-2}{x+2}\right)+\frac{1}{2.4^7}{\left(\frac{x-2}{x+2}\right)}^2+C$$

Commented by msup trace by abdo last updated on 12/Feb/20

$$errorofcalculusfromline6$$$$I=\frac{1}{4^7}\int \frac{{(t-1)}^6}{t^3}dt$$$$=\frac{1}{4^7}\int \frac{\sum _{k=0}^6{C}_6^k{t}^k{(-1)}^{6-k}}{t^3}dt$$$$=\frac{1}{4^7}\sum _{k=0}^6{C}_6^k{(-1)}^{6-k}\int t^{k-3}dt$$$$=.....$$

Answered by MJS last updated on 12/Feb/20

$$\int \frac{dx}{{(x+2)}^5{(x-2)}^3}=?$$$${\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}$$$$\int \frac{P\left(x\right)}{Q\left(x\right)}dx=\frac{P_1\left(x\right)}{Q_1\left(x\right)}+\int \frac{P_2\left(x\right)}{Q_2\left(x\right)}dx$$$$Q_1\left(x\right)=\gcd (Q\left(x\right),Q^{\prime }\left(x\right))={(x+2)}^4{(x-2)}^2$$$$Q_2\left(x\right)=\frac{Q\left(x\right)}{Q_1\left(x\right)}=(x+2)(x-2)$$$$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{d}{dx}\left[\frac{P_1\left(x\right)}{Q_1\left(x\right)}\right]+\frac{P_2\left(x\right)}{Q_2\left(x\right)}$$$$\frac{1}{{(x+2)}^5{(x-2)}^3}=\frac{d}{dx}\left[\frac{{ax}^5+{bx}^4+{cx}^3+{dx}^2+ex+f}{{(x+2)}^4{(x-2)}^2}\right]+\frac{gx+h}{(x+2)(x-2)}$$$$\Rightarrow$$$$P_1\left(x\right)=\frac{15}{4096}{x}^5+\frac{15}{1024}{x}^4-\frac{5}{512}{x}^3-\frac{25}{256}{x}^2-\frac{17}{256}x+\frac{1}{8}$$$$P_2\left(x\right)=\frac{15}{4096}$$$$\int \frac{dx}{{(x+2)}^5{(x-2)}^3}=$$$$=\frac{15x^5+60x^4-40x^3-400x^2-272x+512}{4096{(x+2)}^4{(x-2)}^2}+\frac{15}{4096}\int \frac{dx}{(x+2)(x-2)}=$$$$=\frac{15x^5+60x^4-40x^3-400x^2-272x+512}{4096{(x+2)}^4{(x-2)}^2}+\frac{15}{16384}\ln \mid \frac{x-2}{x+2}\mid +C$$

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