How do you use partial fractions to find the integral #int (x^3-x+3)/(x^2+x-2)dx#?

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Answer 1 · 2016-12-02T08:26:59

THe answer is #=x^2/2-x+ln(∣x-1∣)+ln(∣x+2∣)+C#

The denominator is

#x^2+x-2=(x-1)(x+2)#

But before, let's do a long division

#color(white)(aaaa)##x^3-x+3##color(white)(aaaa)##∣##x^2+x-2#

#color(white)(aaaa)##x^3+x^2-2x##color(white)(aa)##∣##x-1#

#color(white)(aaaa)##0-x^2+x+3#

#color(white)(aaaaaa)##-x^2-x+2#

#color(white)(aaaaaaaa)##0+2x+1#

So,

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

Now we do the decomposition in partial fractions

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

#=A/(x-1)+B/(x+2)=(A(x+2)+B(x-1))/((x-1)(x+2))#

so,

#2x+1=A(x+2)+B(x-1))#

Let #x=1#, #=>#, #3=3A#, #=>#, #A=1#

Let #x=-2#, #=>#, #-3=-3B#, #=>#, #B=1#

So,

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

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

#=x^2/2-x+ln(∣x-1∣)+ln(∣x+2∣)+C#