We start with partial fraction procedure with the unknowns A, B, C
#(3-x)/((x^2+3)(x+3))=(Ax+B)/(x^2+3)+C/(x+3)#
LCD#=(x^2+3)(x+3)#
#(3-x)/((x^2+3)(x+3))=(Ax^2+Bx+3Ax+3B+Cx^2+3C)/((x^2+3)(x+3))#
Transform the numerator so that the numerical coefficients match for both left and right side of the equation
#(0*x^2+(-1)x+3x^0)/((x^2+3)(x+3))=(Ax^2+Cx^2+3Ax+Bx+3B+3C)/((x^2+3)(x+3))#
#(0*x^2+(-1)x+3x^0)/((x^2+3)(x+3))=((A+C)x^2+(3A+B)x+(3B+3C)x^0)/((x^2+3)(x+3))#
We now have the equations to solve for the values of A,B, C
#A+C=0#
#3A+B=-1#
#3B+3C=3#
Simultaneous solution of these equations result to
#A=-1/2# and #B=1/2# and #C=1/2#
#(3-x)/((x^2+3)(x+3))=(-1/2x+1/2)/(x^2+3)+(1/2)/(x+3)#
We can now integrate
#int(3-x)/((x^2+3)(x+3))dx=int [(-1/2x+1/2)/(x^2+3)+(1/2)/(x+3)] dx#
#int(3-x)/((x^2+3)(x+3))dx=int (-1/2x)/(x^2+3)dx+int +(1/2)/(x^2+3)dx+int(1/2)/(x+3) dx#
#int(3-x)/((x^2+3)(x+3))dx=-1/4ln (x^2+3)+(sqrt3)/6 tan^-1 ((sqrt3*x)/3)+1/2ln(x+3)+C_0#
God bless....I hope the explanation is useful.
