Expand the fraction:
#(3x^2 + 3x + 12)/((x - 5)(x^2 + 9)) = A/(x - 5) + (Bx + C)/(x^2 + 9)#
Multiply both sides by the left side denominator:
#3x^2 + 3x + 12 = A(x^2 + 9) + (Bx + C)(x - 5)#
Let x = 5 to make B and C disapper
#3(5)^2 + 3(5) + 12 = A(5^2 + 9)#
#102 = A(34)#
#A = 3#
Substitute 3 for A:
#3x^2 + 3x + 12 = 3(x^2 + 9) + (Bx + C)(x - 5)#
Let x = 0 to make B disappear:
#12 = 3(9) + C(-5)#
#-15 = -5C#
#C = 3#
Substitute 3 for C:
#3x^2 + 3x + 12 = 3(x^2 + 9) + (Bx + 3)(x - 5)#
Let x = 1:
#3 + 3 + 12 = 3(1 + 9) + (B + 3)(1 - 5)#
#18 = 30 -4B - 12#
#B = 0#
#int((3x^2 + 3x + 12)/((x - 5)(x^2 + 9)))dx = 3int1/(x - 5)dx + 3int1/(x^2 + 5)dx#
#int((3x^2 + 3x + 12)/((x - 5)(x^2 + 9)))dx = 3ln|x - 5| + 3sqrt(5)/5tan^-1((sqrt(5))x /5) + C#