1. ## integral

Hi everyone,

Can you help me with this problem, please?

Find the integral

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

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

2. $\int\frac{x^2-x+6}{x^3+3x}\,dx$

You wanna integrate this?

3. Yes, that's what I want to integrate

4. Originally Posted by chocolatelover

$\intx^2-x+6$/x^3+3x= $\intx^2-x+6/x(x^2+3)dx$
you had math tags all over the place here, which is why you got an error. please type what you want to be in one line in only a pair of math tags, no more

5. This is what I'm trying to integrate.

6. $\displaystyle\frac{x^2-x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$
Find $A,B,C$, then integrate.
Can you do it?

7. 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.

8. $\frac{{x^2 - x + 6}}
{{x\left( {x^2 + 3} \right)}} = \frac{a}
{x} + \frac{{bx + c}}
{{x^2 + 3}}$

So this yields

$x^2 - x + 6 = ax^2 + 3a + bx^2 + cx = (a + b)x^2 + cx + 3a$

When $x=0\implies a=2$; it's easy to check that $c=-1\,\therefore\,b=-1$, which yields $\frac{{x^2 - x + 6}}
{{x\left( {x^2 + 3} \right)}} = \frac{2}
{x} - \frac{{x + 1}}
{{x^2 + 3}}$

So... can you take it from there?

9. Hi,

Can you explain to me how you got x=0 and a=2?

Thank you

10. 'cause, when $x=0$, we have that

$6=3a\implies a=2$

You gotta plug $x=0$ into

$x^2 - x + 6 = (a + b)x^2 + cx + 3a$

Got it?

11. Yes, now I understand it.

Then you get integ. 2/xdx-1-x/x sq. + 3xdx=

I know that the first part is "2ln(x)." But what you do with the other part? Do I need to do integration of parts?

Thank you

12. Originally Posted by Krizalid
$\frac{{x^2 - x + 6}}
{{x\left( {x^2 + 3} \right)}} = \frac{2}
{x} - \frac{{x + 1}}
{{x^2 + 3}}$
$\int\frac{x+1}{x^2+3}\,dx=\int\frac x{x^2+3}\,dx+\int\frac1{x^2+3}\,dx$

The first one is easy to take it with a simple substitution, and the second one it's an arctangent.

13. I got ln(x) for the first one, but I not sure what to do for the second one. Can you give me a hint, please?

Thank you

14. Originally Posted by chocolatelover
I got ln(x) for the first one, but I not sure what to do for the second one. Can you give me a hint, please?

Thank you
the first what? integral? the one that was $\int \frac {2}{x}~dx$ ? if so, $\ln x$ is wrong

state specifically which integral you need help with, as far as i can see, there are 3 you have to deal with here

15. For the int. 2/x I got 2ln(x), for the int. x/x sq.+3 I got ln(x) and I'm not sure what to do for the int. 1/x sq. +3

Can you give me a hint or show me what to do, please?

Thank you

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