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I-x-3-2x-2-7x-1-x-3-3-x-2-2-dx-

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Question Number 86638 by M±th+et£s last updated on 29/Mar/20
I=∫((x^3 −2x^2 +7x−1)/((x−3)^3 (x−2)^2 ))dx
I=∫x3−2x2+7x−1(x−3)3(x−2)2dx
Answered by MJS last updated on 29/Mar/20
((x^3 −2x^2 +7x−1)/((x−3)^3 (x−2)^2 ))= =((29)/((x−3)^3 ))−((36)/((x−3)^2 ))+((50)/(x−3))−((13)/((x−2)^2 ))−((50)/(x−2)) ⇒ I=((98x^2 +545x+724)/(2(x−3)^2 (x−2)))+50ln ∣((x−3)/(x−2))∣ +C or t=((x−3)/(x−2)) → dx=(x−2)^2 dt ⇒ I=−13∫dt+50∫(dt/t)−65∫(dt/t^2 )+29∫(dt/t^3 ) and now it′s easy
x3−2x2+7x−1(x−3)3(x−2)2==29(x−3)3−36(x−3)2+50x−3−13(x−2)2−50x−2⇒I=98x2+545x+7242(x−3)2(x−2)+50ln∣x−3x−2∣+Cort=x−3x−2→dx=(x−2)2dt⇒I=−13∫dt+50∫dtt−65∫dtt2+29∫dtt3andnowit′seasy

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