How do you integrate #int (x³ - 7x - 3)/(x² - x -2) # using partial fractions?

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Answer 1 · 2017-02-04T16:19:35

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

As the degree of the numerator is #># degree of the denominator, we have to perform a long division

#color(white)(aaaa)##x^3##color(white)(aaaa)##-7x-3##color(white)(aaaa)##|##x^2-x-2#

#color(white)(aaaa)##x^3##-x^2##-2x##color(white)(aaaaaaaa)##|##x+1#

#color(white)(aaaaa)##0##+x^2##-5x-3#

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

#color(white)(aaaaaaa)####+0##-4x-1#

Also,

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

Therefore,

#(x^3-7x-3)/(x^2-x-2)=(x+1)+(-4x-1)/((x-2)(x+1))#

We can perform the decomposition into partial fractions

#(-4x-1)/((x-2)(x+1))=A/(x-2)+B/(x+1)#

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

We compare the numerators

#-4x-1=A(x+1)+B(x-2)#

Let #x=2#, #=>#, #-9=3A#, #=>#, #A=-3#

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

So,

#(4x+1)/((x-2)(x+1))=-3/(x-2)-1/(x+1)#

Therefore,

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

#int((x^3-7x-3)dx)/(x^2-x-2)=intxdx+int1dx-3intdx/(x-2)-intdx/(x+1)#

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