How do you use partial fraction decomposition to decompose the fraction to integrate #1/((x+2)(x^2+4))#?

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Answer 1 · 2015-10-24T01:11:49

#I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +C#

#1/((x+2)(x^2+4)) = A/(x+2)+(Bx+C)/(x^2+4) =#

#= (A(x^2+4))/((x+2)(x^2+4))+((Bx+C)(x+2))/((x^2+4)(x+2)) =#

#= (Ax^2+4A+Bx^2+2Bx+Cx+2C)/((x+2)(x^2+4)) = #

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

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

#A+B=0 => A=-B#
#2B+C=0 => C=-2B#
#4A+2C=1 => 4(-B)+2(-2B)=1#

#4(-B)+2(-2B)=1#

#-8B=1 => B=-1/8#

#A=-B => A=1/8#

#C=-2B => C=1/4#

#I = int 1/((x+2)(x^2+4)) dx = 1/8int 1/(x+2)dx + int (-1/8x+1/4)/(x^2+4)dx#

#I = 1/8ln|x+2| - 1/8 int (xdx)/(x^2+4) + 1/4int dx/(x^2+4)#

#I = 1/8ln|x+2| - 1/16 int ((x^2+4)'dx)/(x^2+4) + 1/4 *1/2arctan(x/2)#

#I = 1/8ln|x+2| - 1/16 int (d(x^2+4))/(x^2+4) + 1/8arctan(x/2)#

#I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +C#