1) I decomposed integrand into basic fractions
2) I solved basic integrals for finding indefinitely integral.
3) I imposed #f(-1)=0# condition for finding #C#.
I decompose integrand into basic fractions,
#1/[(x+3)*(x^2+9)]=A/(x+3)+(Bx+C)/(x^2+9)#
A(x^2+9)+(Bx+C)(x+3)=1#
Set #x=-3#, #18A=1# or #A=1/18#
Set #x=0#, #9A+3C=1#, hence #C=1/6#
Set #x=1#, #10A+4B+4C=1#, so #B=-1/18#
Consequently,
#int dx/[(x+3)*(x^2+9)]#
=#1/18##int dx/(x+3)#-#1/18##int ((x-3)*dx)/(x^2+9)#
=#1/18##Ln(x+3)#-#1/36##int ((2x-6)*dx)/(x^2+9)#
=#1/18##Ln(x+3)#-#1/36##int (2x*dx)/(x^2+9)#+#1/18#*#int (3*dx)/(x^2+9)#
=#1/18*Ln(x+3)-1/36*Ln(x^2+9)+1/18*arctan(x/3)+C#
After imposing #f(-1)=0# condition,
=#1/18*Ln2-1/36*Ln10+1/18*arctan(-1/3)+C=0#
C=#-1/18*arctan(-1/3)+1/36*Ln10-1/18*Ln2#
#=-1/18*arctan(-1/3)+1/36*Ln(5/2)#
Thus solution of this problem,
#1/18*Ln(x+3)-1/36*Ln(x^2+9)+1/18*arctan(x/3)-1/18*arctan(-1/3)+1/36*Ln(5/2)#
