How do you find int x/((x+2)(x+9))dx∫x(x+2)(x+9)dx using partial fractions?

1 Answer
Harry C.
Nov 3, 2015

9/7ln(x+9) - 2/7ln(x+2)+97ln(x+9)−27ln(x+2)+constant

Explanation:

The equation to be integrated can be written as:

x/((x+2)(x+9))=A/(x+2)+B/(x+9)=(A(x+9)+B(x+2))/((x+2)(x+9)x(x+2)(x+9)=Ax+2+Bx+9=A(x+9)+B(x+2)(x+2)(x+9)

Equating coefficients

x->A(x+9)+B(x+2)=(A+B)x+(9A+2B)x→A(x+9)+B(x+2)=(A+B)x+(9A+2B)

x=(A+B)xx=(A+B)x

Therefore, dividing by xx:

  • A+B=1A+B=1 (1)(1)

Additionally,

  • 9A+2B=09A+2B=0 (2)(2)

Simultaneous Equations

(2)-2times(1)(2)−2×(1)

9A+2B=09A+2B=0 (2)(2)
-2A-2B=-2−2A−2B=−2 (1)(1)

Cancel out the BB terms:

7A=-27A=−2

A=-2/7A=−27

Subbing AA back into (1)(1):

B=9/7B=97

Integrating

Subbing AA and BB back into the original equation:

int9/(7(x+9))-2/(7(x+2))dx∫97(x+9)−27(x+2)dx

=int9/(7(x+9))dx-int2/(7(x+2))dx=∫97(x+9)dx−∫27(x+2)dx

=9/7int1/(x+9)dx-2/7int1/(x+2)dx=97∫1x+9dx−27∫1x+2dx

=9/7ln(x+9) +=97ln(x+9)+constant -2/7ln(x+2)+−27ln(x+2)+constant

=9/7ln(x+9) - 2/7ln(x+2)+=97ln(x+9)−27ln(x+2)+constant

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