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

2 Answers
May 7, 2018

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

Explanation:

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

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

Explanation:

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

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