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

∫(dx/((x−2)^3 (x+1)^2 ))=?

$$\int\frac{{dx}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }=? \\ $$

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

let I =∫ (dx/((x−2)^3 (x+1)^2 )) changement x−2 =t give I =∫ (dt/(t^3 (t+3)^2 )) let decompose F(t) =(1/(t^3 (t+3)^2 )) ⇒ F(t) =(a/t) +(b/t^2 ) +(c/t^3 ) +(d/(t+3)) +(e/((t+3)^2 )) c =lim_(t→0) t^3 F(t) =(1/9) e =lim_(t→−3) (t+3)^2 F(t) =−(1/(27)) ⇒F(t) =(a/t) +(b/t^2 ) +(1/(9t^3 )) +(d/(t+3)) −(1/(27(t+3)^2 )) lim_(t→+∞) tF(t) =0 =a +d ⇒d =−a ⇒ F(t) =(a/t) +(b/t^2 ) +(1/(9t^3 )) −(a/(t+3)) −(1/(27(t+3)^2 )) F(1) =(1/(16)) =a+b +(1/9) −(a/4) −(1/(27.16)) ⇒1 =16a+16b+((16)/9)−4a−(1/(27)) ⇒12a+16b +((47)/(27)) =1 ⇒12a+16b =1−((47)/(27)) =−((20)/(27)) ⇒3a+4b =−(5/(27)) F(−2) =−(1/8) =−(a/2) +(b/4) −(1/(9.8)) −a −(1/(27)) ⇒ (1/8) =(a/2)−(b/4) +(1/(9.8)) +a +(1/(27)) ⇒1 =4a−2b +(1/9) +8a +(8/(27)) ⇒ 12a −2b +((11)/(27)) =1 ⇒12a−2b =1−((11)/(27)) =((16)/(27)) ⇒6a−b =(4/(27)) ⇒ b =6a−(4/(27)) ⇒3a+4(6a−(4/(27)))=−(5/(27)) ⇒27a =((16)/(27))−(5/(27)) =((11)/(27)) ⇒a =((11)/(27^2 )) b =((66)/(27^2 )) −((4.27)/(27^2 )) =−((42)/(27^2 )) ⇒F(t) =((11)/(27^2 t))−((42)/(27^2 t^2 )) +(1/(9t^3 )) −((11)/(27^2 (t+3)))−(1/(27(t+3)^2 )) ⇒ I =∫ F(t)dt =((11)/(27^2 ))ln∣t∣+((42)/(27t))−(1/(18t^2 )) −((11)/(27^2 ))ln∣t+3∣ +(1/(27(t+3))) +C =((11)/(27^2 ))ln∣x−2∣+((42)/(27(x−2))) −(1/(18(x−2)^2 )) −((11)/(27^2 ))ln∣x+1∣+(1/(27(x+1))) +C .

$${let}\:{I}\:=\int\:\:\:\frac{{dx}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{changement}\:{x}−\mathrm{2}\:={t}\:{give} \\ $$$${I}\:=\int\:\:\frac{{dt}}{{t}^{\mathrm{3}} \left({t}+\mathrm{3}\right)^{\mathrm{2}} }\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\frac{\mathrm{1}}{{t}^{\mathrm{3}} \left({t}+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{a}}{{t}}\:+\frac{{b}}{{t}^{\mathrm{2}} }\:+\frac{{c}}{{t}^{\mathrm{3}} }\:+\frac{{d}}{{t}+\mathrm{3}}\:+\frac{{e}}{\left({t}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${c}\:={lim}_{{t}\rightarrow\mathrm{0}} \:{t}^{\mathrm{3}} \:{F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${e}\:={lim}_{{t}\rightarrow−\mathrm{3}} \left({t}+\mathrm{3}\right)^{\mathrm{2}} {F}\left({t}\right)\:=−\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow{F}\left({t}\right)\:=\frac{{a}}{{t}}\:+\frac{{b}}{{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{9}{t}^{\mathrm{3}} }\:+\frac{{d}}{{t}+\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{27}\left({t}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)\:=\mathrm{0}\:={a}\:+{d}\:\Rightarrow{d}\:=−{a}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{a}}{{t}}\:+\frac{{b}}{{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{9}{t}^{\mathrm{3}} }\:−\frac{{a}}{{t}+\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{27}\left({t}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{16}}\:={a}+{b}\:+\frac{\mathrm{1}}{\mathrm{9}}\:−\frac{{a}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{27}.\mathrm{16}}\:\Rightarrow\mathrm{1}\:=\mathrm{16}{a}+\mathrm{16}{b}+\frac{\mathrm{16}}{\mathrm{9}}−\mathrm{4}{a}−\frac{\mathrm{1}}{\mathrm{27}} \\ $$$$\Rightarrow\mathrm{12}{a}+\mathrm{16}{b}\:\:+\frac{\mathrm{47}}{\mathrm{27}}\:=\mathrm{1}\:\Rightarrow\mathrm{12}{a}+\mathrm{16}{b}\:=\mathrm{1}−\frac{\mathrm{47}}{\mathrm{27}}\:=−\frac{\mathrm{20}}{\mathrm{27}}\:\Rightarrow\mathrm{3}{a}+\mathrm{4}{b}\:=−\frac{\mathrm{5}}{\mathrm{27}} \\ $$$${F}\left(−\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{8}}\:=−\frac{{a}}{\mathrm{2}}\:+\frac{{b}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{9}.\mathrm{8}}\:−{a}\:−\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\:=\frac{{a}}{\mathrm{2}}−\frac{{b}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{9}.\mathrm{8}}\:+{a}\:+\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow\mathrm{1}\:=\mathrm{4}{a}−\mathrm{2}{b}\:+\frac{\mathrm{1}}{\mathrm{9}}\:+\mathrm{8}{a}\:+\frac{\mathrm{8}}{\mathrm{27}}\:\Rightarrow \\ $$$$\mathrm{12}{a}\:−\mathrm{2}{b}\:+\frac{\mathrm{11}}{\mathrm{27}}\:=\mathrm{1}\:\Rightarrow\mathrm{12}{a}−\mathrm{2}{b}\:=\mathrm{1}−\frac{\mathrm{11}}{\mathrm{27}}\:=\frac{\mathrm{16}}{\mathrm{27}}\:\Rightarrow\mathrm{6}{a}−{b}\:=\frac{\mathrm{4}}{\mathrm{27}}\:\Rightarrow \\ $$$${b}\:=\mathrm{6}{a}−\frac{\mathrm{4}}{\mathrm{27}}\:\Rightarrow\mathrm{3}{a}+\mathrm{4}\left(\mathrm{6}{a}−\frac{\mathrm{4}}{\mathrm{27}}\right)=−\frac{\mathrm{5}}{\mathrm{27}}\:\Rightarrow\mathrm{27}{a}\:=\frac{\mathrm{16}}{\mathrm{27}}−\frac{\mathrm{5}}{\mathrm{27}}\:=\frac{\mathrm{11}}{\mathrm{27}}\:\Rightarrow{a}\:=\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} } \\ $$$${b}\:=\frac{\mathrm{66}}{\mathrm{27}^{\mathrm{2}} }\:−\frac{\mathrm{4}.\mathrm{27}}{\mathrm{27}^{\mathrm{2}} }\:=−\frac{\mathrm{42}}{\mathrm{27}^{\mathrm{2}} }\:\Rightarrow{F}\left({t}\right)\:=\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} {t}}−\frac{\mathrm{42}}{\mathrm{27}^{\mathrm{2}} {t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{9}{t}^{\mathrm{3}} }\:−\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} \left({t}+\mathrm{3}\right)}−\frac{\mathrm{1}}{\mathrm{27}\left({t}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{I}\:=\int\:{F}\left({t}\right){dt}\:=\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} }{ln}\mid{t}\mid+\frac{\mathrm{42}}{\mathrm{27}{t}}−\frac{\mathrm{1}}{\mathrm{18}{t}^{\mathrm{2}} }\:−\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} }{ln}\mid{t}+\mathrm{3}\mid\:+\frac{\mathrm{1}}{\mathrm{27}\left({t}+\mathrm{3}\right)}\:+{C} \\ $$$$=\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} }{ln}\mid{x}−\mathrm{2}\mid+\frac{\mathrm{42}}{\mathrm{27}\left({x}−\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\mathrm{18}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:−\frac{\mathrm{11}}{\mathrm{27}^{\mathrm{2}} }{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{27}\left({x}+\mathrm{1}\right)}\:+{C}\:. \\ $$$$ \\ $$

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

thanks sir

$${thanks}\:{sir} \\ $$

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

you are welcome sir.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

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