#int(2x+4)/((x+2)(x-6)(x²+3))dx#
#=int(2(cancel(x+2)))/((cancel(x+2))(x-6)(x²+3))dx#
#=2int1/((x-6)(x²+3))dx#
Now let #1/((x-6)(x²+3))=A/(x-6)+(Bx+C)/(x²+3)#
#1/(cancel((x-6)(x²+3)))=(Ax²+3A+Bx²-6Bx+Cx-6C)/(cancel((x-6)(x²+3)))#
#(A+B)x²+(C-6B)x+3A-6C=1#
By identification :
#A+B=0#[1]
#C-6B=0#[2]
#3A-6C=1#[3]
#A+B=0#[1]
#6A+C=0#[2]
#6A-12C=2#[3]
#A+B=0#[1]
#6A+C=0#[2]
#-13C=2#[3]
#A+B=0#[1]
#6A-2/13=0#[2]
#C=-2/13#[3]
#B=-1/39#[1]
#A=1/39#[2]
#C=-2/13#[3]
So:
#int(2x+4)/((x+2)(x-6)(x²+3))dx=int(1/(39(x-6))+((-1/39)x-2/13)/(x²+3))dx#
#=int(1/39(1/(x-2/13))dx-int(1/39x+2/13)/(x²+3)dx#
#=1/39int1/(x-2/13)dx-1/78int(2x/(x²+3))dx-1/39int(6/(x²+3))dx#
#=1/39ln(x-2/13)-1/78ln(x²+3)-2/13int(1/(x²+3))dx#
#=1/39ln(x-2/13)-1/78ln(x²+3)-2/13*1/sqrt6arctan(x/sqrt3)#
#=1/39ln(x-2/13)-1/78ln(x²+3)-2/(13sqrt6)arctan(x/sqrt6)+C#,#C in RR#
\0/ here's our answer!
