Hi everyone,

Can you help me with this problem, please?

Find the integral

$\displaystyle \intx^2-x+6$/x^3+3x=$\displaystyle \intx^2-x+6/x(x^2+3)dx$

int. x^2-x+6/x^3+3xdx=int. x^2-x+6/x(x^2+3)dx

=A/x+bx+c/x^2+3

=A(x^2+3)+Bx^2+cx/x(x^2+3)=x^2-x+6

Ax^2+3A+Bx^2+Cx=x^2-x+6

Ax^2+Bx^2=x^2

x^2(a+B)=x^2

A+b=1

A=1-B

3A=-x

cx=6

I don't know what to do from here. I was going to plug in 3a with 3(1-B), but that isn't going to help since I still have an x. Can someone give me a hint, please?

Thank you