Question #186224

Integration by Partial Fractions


1.) ∫(3x^2-x+1)/(x^3-x^2)dx


2.) ∫(t^2dt)/(t+1)^3


1
Expert's answer
2021-05-07T11:52:40-0400

(1)

(3x2x+1)(x3x2)dx=3x22x+x1+2x3x2dx=3x22xx3x2dx+x1x2(x1)dx+2x2(x1)dx∫\dfrac{(3x^2-x+1)}{(x^3-x^2)}dx\\[9pt]=\dfrac{3x^2-2x+x-1+2}{x^3-x^2}dx\\[9pt]=\dfrac{3x^2-2x}{x^3-x^2}dx+\dfrac{x-1}{x^2(x-1)}dx+\dfrac{2}{x^2(x-1)}dx\\[9pt]


Let x3x2=tx^3-x^2=t

(3x22x)dx=dt\Rightarrow (3x^2-2x)dx=dt and using partial fraction in the third term-



=dtt+1x2dx+2x1dx2xdx2x2dx=lnt1x+2ln(x1)2lnx+2x+C=ln(x3x2)+1x+2lnx1x+C=\dfrac{dt}{t}+\dfrac{1}{x^2}dx+\dfrac{2}{x-1}dx-\dfrac{2}{x}dx-\dfrac{2}{x^2}dx\\[9pt]=lnt-\dfrac{1}{x}+2ln(x-1)-2lnx+\dfrac{2}{x}+C\\[9pt]=ln(x^3-x^2)+\dfrac{1}{x}+2ln\dfrac{x-1}{x}+C


(2)

(t2)(t+1)3dt=(t+11)2(t+1)3dt=(t+1)2+12(t+1)(t+1)3dt=1t+1dt+1(t+1)3dt2(t+1)2dt=ln(t+1)12(t+1)2+2(t+1)+C∫\dfrac{(t^2)}{(t+1)^3}dt\\[9pt]=\dfrac{(t+1-1)^2}{(t+1)^3}dt\\[9pt]=\dfrac{(t+1)^2+1-2(t+1)}{(t+1)^3}dt\\[9pt]=\dfrac{1}{t+1}dt+\dfrac{1}{(t+1)^3}dt-\dfrac{2}{(t+1)^2}dt\\[9pt]=ln(t+1)-\dfrac{1}{2(t+1)^2}+\dfrac{2}{(t+1)}+C


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