How do you integrate #int 1/(x^2-x-20) dx# using partial fractions?

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Answer 1 · 2018-05-07T02:32:52

#1/9ln|x-5|-1/9ln|x+4|+C#

Factor the denominator of the integrand:

#x^2-x-20=(x-5)(x+4)#

Use partial fraction decomposition on the simplified integrand:

#1/((x-5)(x+4))=A/(x-5)+B/(x+4)#

Add up the right side:

#1/((x-5)(x+4))=(A(x+4)+B(x-5))/((x-5)(x+4))#

Equate numerators:

#1=A(x+4)+B(x-5)#

We need to find #A,B.# We can do this by plugging in values of #x# that send one term to #0# and keep the other:

#x=-4:#

#1=-9B, B=-1/9#

#x=5:#

#1=9A, A=1/9#

Thus, our integral becomes

#int(1/9)/(x-5)-(1/9)/(x+4)dx=1/9ln|x-5|-1/9ln|x+4|+C#

Answer 2 · 2018-05-07T02:36:23

#1/9 * lnabs(x-5)- 1/9 * lnabs(x+4)#

#1/((x-5)(x+4)) = 1/(9*(x-5)) - 1/(9*(x+4))#

#int1/((x-5)(x+4)) = int1/(9*(x-5)) - 1/(9*(x+4))dx#

#int1/((x-5)(x+4)) = 1/9 * lnabs(x-5)- 1/9 * lnabs(x+4)#