Integrate the following: (please show work im so confused!!)?

Question

int [(x^2-x+6)/(x^3+3x)]dx

Answer 1

(x^2 - x + 6) / (x^3 + 3x)
= (x^2 - x + 6) / (x(x^2 + 3)): use partial fractions.
= A/x + (Bx + C)/(x^2 + 3)
<=> x^2 - x + 6 = A(x^2 + 3) + (Bx + C)x
= (A+B) x^2 + Cx + 3A
<=> A+B = 1, C = -1, 3A = 6
<=> A = 2, B = -1, C = -1
So ∫(x^2 - x + 6) / (x^3 + 3x) dx
= ∫(2/x) dx + ∫(-x - 1)/(x^2 + 3) dx
= 2 ln |x| + ∫(-du/2u) + ∫(-1/(x^2 + 3) dx using u = x^2 + 3
= 2 ln |x| - (1/2) ln (x^2 + 3) + ∫(-1/(3tan^2 θ + 3)) (√3 sec^2 θ) dθ using x = √3 tan θ
= 2 ln |x| - (1/2) ln (x^2 + 3) + ∫(-1/(3sec^2 θ)) (√3 sec^2 θ) dθ
= 2 ln |x| - (1/2) ln (x^2 + 3) + ∫(-1/√3) dθ
= 2 ln |x| - (1/2) ln (x^2 + 3) - 1/√3 arctan (x/√3) + c.
(x = √3 tan θ => θ = arctan (x/√3).)

Answer 2

int [(x^2-x+6)/(x^3+3x)]dx

um lets see.
i get an ans.
but im too frustrated to type it in the computer.

Answer 3

scarlet is correct.