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

1 Answer

#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)]#

=#1/18##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)#