How do you find #int (x-1)/((x+2)(x^2+3))dx# using partial fractions?

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Answer 1 · 2018-04-02T09:19:43

The answer is #=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/(7sqrt3)arctan(x/sqrt3)+C#

Perform the decomposition into partial fractions

#(x-1)/((x+2)(x^2+3))=A/(x+2)+(Bx+C)/(x^2+3)#

#=(A(x^2+3)+(Bx+C)(x+2))/((x+2)(x^2+3))#

The denominators are the same, compare the numerators

#x-1=A(x^2+3)+(Bx+C)(x+2)#

Let #x=-2#, #=>#, #-3=7A#, #=>#, #A=-3/7#

Let #x=0#, #=>#, #-1=3A+2C#

#2C=-1-3A=-1+9/7=2/7#

#C=1/7#

Coefficients of #x^2#

#0=A+B#

#B=-A=3/7#

Therefore,

#(x-1)/((x+2)(x^2+3))=(-3/7)/(x+2)+((3/7)x+(1/7))/(x^2+3)#

So,

#int((x-1)dx)/((x+2)(x^2+3))=int(-3/7dx)/(x+2)+int(((3/7)x+(1/7))dx)/(x^2+3)#

#=-3/7ln(|x+2|)+3/14int(2xdx)/(x^2+3)+1/7int(1dx)/(x^2+3)#

#=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/21int(dx)/((x/sqrt3)^2+1)#

#=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/21*sqrt3arctan(x/sqrt3)+C#

#=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/(7sqrt3)arctan(x/sqrt3)+C#