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

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Answer 1 · 2015-11-03T02:31:03

#9/7ln(x+9) - 2/7ln(x+2)+#constant

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)#

Equating coefficients

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

#x=(A+B)x#

Therefore, dividing by #x#:

Additionally,

Simultaneous Equations

#(2)-2times(1)#

#9A+2B=0# #(2)#
#-2A-2B=-2# #(1)#

Cancel out the #B# terms:

#7A=-2#

#A=-2/7#

Subbing #A# back into #(1)#:

#B=9/7#

Integrating

Subbing #A# and #B# back into the original equation:

#int9/(7(x+9))-2/(7(x+2))dx#

#=int9/(7(x+9))dx-int2/(7(x+2))dx#

#=9/7int1/(x+9)dx-2/7int1/(x+2)dx#

#=9/7ln(x+9) +#constant #-2/7ln(x+2)+#constant

#=9/7ln(x+9) - 2/7ln(x+2)+#constant