How do you integrate #(x - 16) / (x^2 + x - 2)# using partial fractions?

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1 Answer
Steve M
Oct 9, 2016

#int((x-16)/(x^2+x-2))dx = 6 ln|x+2|-5ln|x-1|+c#

Explanation:

We first expand the given expression into partial fractions:

#(x-16)/(x^2+x-2)-=(x-16)/((x+2)(x-1))#
#(x-16)/(x^2+x-2)-=A/(x+2)+B/(x-1)#
#(x-16)/(x^2+x-2)-=(A(x-1)+B(x+2))/((x+2)(x-1))#

And so. #(x-16)-=A(x-1)+B(x+2)#

Put #x=1=>1-16=0+B(3)#
#:. 3B=-15=>B=-5#

Put #x=-2=>-2-16=A(-2-1)+0#
#:. -3A=-18=>A=6#

So the partial fraction decomposition is:
#(x-16)/(x^2+x-2)-=6/(x+2)-5/(x-1)#

We now want to integrate; so
#int((x-16)/(x^2+x-2))dx = int(6/(x+2)-5/(x-1))dx#
#int((x-16)/(x^2+x-2))dx = 6 int(1/(x+2))dx-5int(1/(x-1))dx#
#int((x-16)/(x^2+x-2))dx = 6 ln|x+2|-5ln|x-1|+c#

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