How do you integrate #int (1-2x^2)/((x-2)(x+6)(x+4)) # using partial fractions?

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Answer 1 · 2016-01-12T10:20:38

#-7/48ln (x-2)-71/16ln(x+6)+31/12ln(x+4) + C#

from the given
#int (1-2x^2)/((x-2)(x+6)(x+4)) # #dx#

#A/(x-2)+B/(x+6)+C/(x+4)# = #(1-2x^2)/((x-2)(x+6)(x+4)) #

then

#A(x+6)(x+4)+B(x-2)(x+4)+C(x-2)(x+6)#

= #1 - 2x^2#

then

#A(x^2+10x+24)+B(x^2+2x-8)+C(x^2+4x-12)#

=#1-2x^2#

The 3 equations are now defined

#A+B+C=-2#

#10A+2B+4C=0#

#24A-8B-12C=1#

Use Algebra to find A, B, C

#A=-7/48#

#B=-71/16#

#C=31/12#

The integral becomes:

#int (-7/48)/(x-2)##dx#+#int (-71/16)/(x+6)##dx#+#int (31/12)/(x+4)##dx#

# -7/48ln(x-2)# -#71/16ln(x+6)#+# 31/12ln(x+4)#