How do you integrate #int (3-x)/((x^2+3)(x+3)) dx# using partial fractions?

Question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-06T00:20:25

#I= -1/4ln(x^2+3)+1/(2sqrt3)arctan(x/sqrt3) +1/2ln|x+3| +C#

#(3-x)/((x^2+3)(x+3))=(Ax+B)/(x^2+3)+C/(x+3)=#

#=((Ax+B)(x+3)+C(x^2+3))/((x^2+3)(x+3))=#

#=(Ax^2+Bx+3Ax+3B+Cx^2+3C)/((x^2+3)(x+3))=#

#=((A+C)x^2+(B+3A)x+(3B+3C))/((x^2+3)(x+3))#

#A+C=0 => A=-C#
#B+3A=-1 => B=-1+3C#
#3B+3C=3 => B+C=1#

#-1+3C+C=1 => 4C=2 => C=1/2#

#A=-1/2#

#B=1-C=1/2#

#I=int (3-x)/((x^2+3)(x+3))dx = int (-1/2x+1/2)/(x^2+3)dx + 1/2intdx/(x+3)#

#I=-1/4int(2xdx)/(x^2+3)+1/2intdx/(x^2+(sqrt3)^2)+1/2ln|x+3|#

#I= -1/4ln(x^2+3)+1/(2sqrt3)arctan(x/sqrt3) +1/2ln|x+3| +C#