Question

Integrate (x^2-x+6)/(x^3+3x)

Answer 1

have you tried using rational expressions?

Answer 2

I believe it's meant to be done by partial fractions but I am having a hard time of it

Answer 3

x^2-x+6 = a(x)+b(x^2+3) is as far as I have got.

Answer 4

x(x^2+ 3)

so partial fraction is:
A/x + (Bx +C)/x^2+3
x^2 -x +6 = Ax^2 + 3A + Bx^2 + Cx

Gaussian method:
coeefficients of x^2:
1 = A + B
coefficients of x
-1 = C
coefficients of constant:
6 = 3A
A = 2
B = -1

so your equation is 2/x + (-1 -x)/x+3

so...
2/x - (1+x)/(x+3)

can you do it from there?

Answer 5

Can I ask why is it (Bx+C)? Instead of just B/x^2+3?

Answer 6

because x^2 + 3 is an irreducible quadratic fractions (Case III of partial fractions)
If you have irreducible quadratic functions, the numerator is composed of Bx + C or any other distinct letters

Answer 7

Thanks again, I can see that part in my book but it's not very well explained. Thanks !