You wanna integrate this?
Hi everyone,
Can you help me with this problem, please?
Find the integral
/x^3+3x=
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
Hi,
This is what I've done so far, but I'm stuck.
=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.