1) I decomposed integrand into basic fractions
2) I solved basic integrals for finding indefinitely integral.
3) I imposed f(-1)=0f(−1)=0 condition for finding CC.
I decompose integrand into basic fractions,
1/[(x+3)*(x^2+9)]=A/(x+3)+(Bx+C)/(x^2+9)1(x+3)⋅(x2+9)=Ax+3+Bx+Cx2+9
A(x^2+9)+(Bx+C)(x+3)=1#
Set x=-3x=−3, 18A=118A=1 or A=1/18A=118
Set x=0x=0, 9A+3C=19A+3C=1, hence C=1/6C=16
Set x=1x=1, 10A+4B+4C=110A+4B+4C=1, so B=-1/18B=−118
Consequently,
int dx/[(x+3)*(x^2+9)]∫dx(x+3)⋅(x2+9)
=1/18118int dx/(x+3)∫dxx+3-1/18118int ((x-3)*dx)/(x^2+9)∫(x−3)⋅dxx2+9
=1/18118Ln(x+3)ln(x+3)-1/36136int ((2x-6)*dx)/(x^2+9)∫(2x−6)⋅dxx2+9
=1/18118Ln(x+3)ln(x+3)-1/36136int (2x*dx)/(x^2+9)∫2x⋅dxx2+9+1/18118*int (3*dx)/(x^2+9)∫3⋅dxx2+9
=1/18*Ln(x+3)-1/36*Ln(x^2+9)+1/18*arctan(x/3)+C118⋅ln(x+3)−136⋅ln(x2+9)+118⋅arctan(x3)+C
After imposing f(-1)=0f(−1)=0 condition,
=1/18*Ln2-1/36*Ln10+1/18*arctan(-1/3)+C=0118⋅ln2−136⋅ln10+118⋅arctan(−13)+C=0
C=-1/18*arctan(-1/3)+1/36*Ln10-1/18*Ln2−118⋅arctan(−13)+136⋅ln10−118⋅ln2
=-1/18*arctan(-1/3)+1/36*Ln(5/2)=−118⋅arctan(−13)+136⋅ln(52)
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)118⋅ln(x+3)−136⋅ln(x2+9)+118⋅arctan(x3)−118⋅arctan(−13)+136⋅ln(52)
