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Question Number 32958 by artibunja last updated on 07/Apr/18

Commented by MJS last updated on 07/Apr/18

((x^2 −3x−10)/(3x^2 −5x+2))=(1/3)+(((−3+(5/3))x+(−10−(2/3)))/(3x^2 −5x+2))= =(1/3)−(4/3)×((x+8)/(3x^2 −5x+2)) 3x^2 −5x+2=(x−1)(3x−2) ((x+8)/(3x^2 −5x+2))=(N_1 /(x−1))+(N_2 /(3x−2))= =((N_1 (3x−2)+N_2 (x−1))/((x−1)(3x−2)))= =(((3N_1 +N_2 )x−(2N_1 +N_2 ))/((x−1)(3x−2))) ⇒ ⇒ 3N_1 +N_2 =1 ∧ −2N_1 −N_2 =8 ⇒ ⇒ N_1 =9 ∧ N_2 =−26 ∫((x^2 −3x−10)/(3x^2 −5x+2))dx= =∫(1/3)−(4/3)((9/(x−1))−((26)/(3x−2)))dx= =∫(1/3)dx−12∫(1/(x−1))dx+((104)/3)∫(1/(3x−2))dx= =(x/3)−12ln(x−1)+((104)/9)ln(3x−2)+C

x2−3x−103x2−5x+2=13+(−3+53)x+(−10−23)3x2−5x+2==13−43×x+83x2−5x+23x2−5x+2=(x−1)(3x−2)x+83x2−5x+2=N1x−1+N23x−2==N1(3x−2)+N2(x−1)(x−1)(3x−2)==(3N1+N2)x−(2N1+N2)(x−1)(3x−2)⇒⇒3N1+N2=1∧−2N1−N2=8⇒⇒N1=9∧N2=−26∫x2−3x−103x2−5x+2dx==∫13−43(9x−1−263x−2)dx==∫13dx−12∫1x−1dx+1043∫13x−2dx==x3−12ln(x−1)+1049ln(3x−2)+C

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