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How do you integrate #int (1-2x^2)/((x-2)(x+7)(x+1)) # using partial fractions?

Answer 1

Cameron Alexander

#-1/9ln|(x-2)|+5/18ln|(x+7)|-1/6ln|(x+1)|+C#

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

Considering the integrand, we have an easy identity because all of the denominators are linear.

#(1-2x)/((x-2)(x+7)(x+1))-+A/(x-2)+B/(x+7)+C/(x+1)#

multiply both sides by#((x-2)(x+7)(x+1))#

after cancelling we end up with

#(1-2x)-=A(x+7)(x+1)+B(x-2)(x+1)+C(x-2)(x+7)#

#"let " x=2#

#1-4=Axx9xx3+0+0#

#-3=27A#

#=>A=-1/9#

#"let "x=-7#

#1-(2xx-7)=0+Bxx-9xx-6+0#

#15=54B#

#=>B=15/54=5/18#

#"let "x=-1#

#1-(2xx-1)=0+0+Cxx(-1-2)(-1+7)#

#=>3=-18C#

#=>C=-1/6#

#:.int(1-2x)/((x-2)(x+7)(x+1))dx#

#int(-1/(9(x-2))+5/(18(x+7))-1/(6(x+1)))dx#

using#color(blue)(int(f'(x))/(f(x))dx=ln|f(x)|+C)#

we have

#-1/9ln|(x-2)|+5/18ln|(x+7)|-1/6ln|(x+1)|+C#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.