How do you integrate #int (x-3x^2)/((x+3)(x-1)(x-7)) # using partial fractions?

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Answer 1 · 2016-09-27T07:43:24

#=-3/4In(x+3)+1/12In(x-1)-7/3In(x-7)#

first ,ignore the integrate sign and do partial fraction of the function

#(x-3x^2)/((x+3)(x-1)(x-7))=A/(x+3)+B/(x-1)+C/(x-7)#

#x-3x^2=A(x-1)(x-7))+B(x+3)(x-7))+C(x+3)(x-1)#

subtitute #x=-3#

#(-3)-3(-3)^2=A(-3-1)(-3-7)#

#-3-27=A(-4)(-10)#

#-30=40A#

#A=-3/4#

subtitute #x=1#

#(1)-3(1)^2=B(1+3)(1-7)#

#-2=B(4)(-6)#

#-2=-24B#

#B=1/12#

subtitute #x=7#

#(7)-3(7)^2=C(7+3)(7-1)#

#7-147=C(10)(6)#

#-140=60C#

#C=-7/3#

#int(x-3x^2)/((x+3)(x-1)(x-7))#

#=int-3/(4(x+3))+1/(12(x-1))-7/(3(x-7))#

#=-3/4int1/(x+3)+1/12int1/(x-1)-7/3int1/(x-7)#

#=-3/4In(x+3)+1/12In(x-1)-7/3In(x-7)#