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Question Number 73479 by abdomathmax last updated on 13/Nov/19
find ∫ ((3x+2)/((x+1)^2 (x−2)^3 ))dx
$${find}\:\int\:\:\:\:\frac{\mathrm{3}{x}+\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} }{dx} \\ $$
Commented by abdomathmax last updated on 17/Nov/19
let decompose the fraction F(x)=((3x+2)/((x+1)^2 (x−2)^3 )) F(x)=(a/(x+1)) +(b/((x+1)^2 )) +(c/(x−2)) +(d/((x−2)^2 )) +(e/((x−2)^3 )) b =(x+1)^2 F(x)∣_(x=−1) =((−1)/(−27)) =(1/(27)) e=(x−2)^3 F(x)∣_(x=2) = (8/(27)) ⇒ F(x)=(a/(x+1))+(1/(27(x+1)^2 )) +(c/(x−2)) +(d/((x−2)^2 )) +(8/(27(x−2)^3 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x) =(a/(x+1)) +(1/(27(x+1)^2 ))−(a/(x−2)) +(d/((x−2)^2 )) +(8/(27(x−2)^3 )) F(0) =−(1/4) =a +(1/(27)) +(a/2) +(d/4) −(1/(27)) ⇒ −(1/4) =(3/2)a +(d/4) ⇒−1=6a+d F(1) =(5/(4(−1))) =−(5/4) =(a/2) +(1/(4×27)) +a+d−(8/(27)) =(3/2)a +d +((1/4)−8)(1/(27)) ⇒ −5 =6a +4d+ (1−32)×(1/(27)) =6a+4d −((31)/(27)) 6a+4d =((31)/(27))−5 =((31−135)/(27)) =((104)/(27)) ⇒ 3a+2d =((52)/(27)) we get the system { ((6a+d=−1)),((3a+2d=((52)/(27)))) :} Δ_s =12−3 =9 a=( determinant (((−1 1)),((((52)/(27 )) 2)))/9) =((−2−((52)/(27)))/9) =((−54−52)/(27×9)) =... b=( determinant (((6 −1)),((3 ((52)/(27)))))/9) =((((6×52)/(27))+3)/9) =... the coefficients are known ⇒ ∫ F(x)dx =aln∣x+1∣−(1/(27(x+1))) −aln∣x−2∣−(d/(x−2)) −(4/(27))(1/((x−2)^2 )) +C.
$${let}\:{decompose}\:{the}\:{fraction}\:{F}\left({x}\right)=\frac{\mathrm{3}{x}+\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{x}−\mathrm{2}}\:+\frac{{d}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{{e}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$${b}\:=\left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\mid_{{x}=−\mathrm{1}} =\frac{−\mathrm{1}}{−\mathrm{27}}\:=\frac{\mathrm{1}}{\mathrm{27}} \\ $$$${e}=\left({x}−\mathrm{2}\right)^{\mathrm{3}} {F}\left({x}\right)\mid_{{x}=\mathrm{2}} \:\:=\:\frac{\mathrm{8}}{\mathrm{27}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{27}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{x}−\mathrm{2}}\:+\frac{{d}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{\mathrm{8}}{\mathrm{27}\left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{27}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{{a}}{{x}−\mathrm{2}}\:+\frac{{d}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{\mathrm{8}}{\mathrm{27}\left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$${F}\left(\mathrm{0}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\:={a}\:+\frac{\mathrm{1}}{\mathrm{27}}\:+\frac{{a}}{\mathrm{2}}\:+\frac{{d}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{2}}{a}\:+\frac{{d}}{\mathrm{4}}\:\Rightarrow−\mathrm{1}=\mathrm{6}{a}+{d} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{5}}{\mathrm{4}\left(−\mathrm{1}\right)}\:=−\frac{\mathrm{5}}{\mathrm{4}}\:=\frac{{a}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}×\mathrm{27}}\:+{a}+{d}−\frac{\mathrm{8}}{\mathrm{27}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{a}\:\:+{d}\:\:+\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{8}\right)\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow \\ $$$$−\mathrm{5}\:=\mathrm{6}{a}\:+\mathrm{4}{d}+\:\left(\mathrm{1}−\mathrm{32}\right)×\frac{\mathrm{1}}{\mathrm{27}}\:=\mathrm{6}{a}+\mathrm{4}{d}\:−\frac{\mathrm{31}}{\mathrm{27}} \\ $$$$\mathrm{6}{a}+\mathrm{4}{d}\:=\frac{\mathrm{31}}{\mathrm{27}}−\mathrm{5}\:=\frac{\mathrm{31}−\mathrm{135}}{\mathrm{27}}\:=\frac{\mathrm{104}}{\mathrm{27}}\:\Rightarrow \\ $$$$\mathrm{3}{a}+\mathrm{2}{d}\:=\frac{\mathrm{52}}{\mathrm{27}}\:\:{we}\:{get}\:{the}\:{system}\:\:\begin{cases}{\mathrm{6}{a}+{d}=−\mathrm{1}}\\{\mathrm{3}{a}+\mathrm{2}{d}=\frac{\mathrm{52}}{\mathrm{27}}}\end{cases} \\ $$$$\Delta_{{s}} =\mathrm{12}−\mathrm{3}\:=\mathrm{9} \\ $$$${a}=\frac{\begin{vmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{52}}{\mathrm{27}\:}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}}{\mathrm{9}}\:=\frac{−\mathrm{2}−\frac{\mathrm{52}}{\mathrm{27}}}{\mathrm{9}}\:=\frac{−\mathrm{54}−\mathrm{52}}{\mathrm{27}×\mathrm{9}}\:=… \\ $$$${b}=\frac{\begin{vmatrix}{\mathrm{6}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{52}}{\mathrm{27}}}\end{vmatrix}}{\mathrm{9}}\:=\frac{\frac{\mathrm{6}×\mathrm{52}}{\mathrm{27}}+\mathrm{3}}{\mathrm{9}}\:=… \\ $$$${the}\:{coefficients}\:{are}\:{known}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:={aln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{27}\left({x}+\mathrm{1}\right)}\:−{aln}\mid{x}−\mathrm{2}\mid−\frac{{d}}{{x}−\mathrm{2}} \\ $$$$−\frac{\mathrm{4}}{\mathrm{27}}\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+{C}. \\ $$$$ \\ $$
Answered by MJS last updated on 13/Nov/19
∫((3x+2)/((x−2)^3 (x+1)^2 ))dx= =(8/9)∫(dx/((x−2)^3 ))−(7/(27))∫(dx/((x−2)^2 ))+(2/(27))∫(dx/(x−2))+(1/(27))∫(dx/((x+1)^2 ))−(2/(27))∫(dx/(x+1))= =−(4/(9(x−2)^2 ))+(7/(27(x−2)))+(2/(27))ln (x−2) −(1/(27(x+1)))−(2/(27))ln (x+1) = =((2x^2 −5x−10)/(9(x−2)^2 (x+1)))+(2/(27))ln ∣((x−2)/(x+1))∣ +C
$$\int\frac{\mathrm{3}{x}+\mathrm{2}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{8}}{\mathrm{9}}\int\frac{{dx}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{7}}{\mathrm{27}}\int\frac{{dx}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{27}}\int\frac{{dx}}{{x}−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{27}}\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{27}}\int\frac{{dx}}{{x}+\mathrm{1}}= \\ $$$$=−\frac{\mathrm{4}}{\mathrm{9}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{7}}{\mathrm{27}\left({x}−\mathrm{2}\right)}+\frac{\mathrm{2}}{\mathrm{27}}\mathrm{ln}\:\left({x}−\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{27}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{27}}\mathrm{ln}\:\left({x}+\mathrm{1}\right)\:= \\ $$$$=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{10}}{\mathrm{9}\left({x}−\mathrm{2}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)}+\frac{\mathrm{2}}{\mathrm{27}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}\mid\:+{C} \\ $$

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