What is f(x) = int 1/((x+3)(x^2+9)) dx if f(-1) = 0 ?

1 Answer
Sep 9, 2017

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)

Explanation:

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)]

=1/18int dx/(x+3)-1/18int ((x-3)*dx)/(x^2+9)

=1/18Ln(x+3)-1/36int ((2x-6)*dx)/(x^2+9)

=1/18Ln(x+3)-1/36int (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)