Find the integral of 4x/ (x^2 - 4 )(x-3) dx ?

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Answer 1 · 2018-08-14T20:25:39

#I=-2/5{5ln|x-2|+ln|x+2|-6ln|x-3|}+c#

Here,
#I=int(4x)/ ((x^2 - 4 )(x-3)) dx=int(4x)/((x-2)(x+2)(x-3))dx#

To obtain partial fractions :

#(4x)/((x-2)(x+2)(x-3))=A/(x-2)+B/(x+2)+C/(x-3)#

#4x=A(x+2)(x-3)+B(x-2)(x-3)+C(x-2)(x+2)#

#:.x=2=>8=A(4)(-1)=>A=-2#

#x=-2=>-8=B(-4)(-5)=>B=-2/5#

#x=3=>12=C(1)(5)=>C=12/5#

#:.I=int[(-2)/(x-2)+(-2/5)/(x+2)+(12/5)/(x-3)]dx#

#:.I=-2/5int[5/(x-2)+1/(x+2)-6/(x-3)]dx#

#:.I=-2/5{5ln|(x-2)|+ln|(x+2)|-6ln|(x-3)|}+c#