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Math Help - integral

  1. #1
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    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
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  2. #2
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    \int\frac{x^2-x+6}{x^3+3x}\,dx

    You wanna integrate this?
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  3. #3
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    Yes, that's what I want to integrate
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post

    \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
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  5. #5
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    This is what I'm trying to integrate.
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  6. #6
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    \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?
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  7. #7
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    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.
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  8. #8
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    \frac{{x^2 - x + 6}}<br />
{{x\left( {x^2 + 3} \right)}} = \frac{a}<br />
{x} + \frac{{bx + c}}<br />
{{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}}<br />
{{x\left( {x^2 + 3} \right)}} = \frac{2}<br />
{x} - \frac{{x + 1}}<br />
{{x^2 + 3}}

    So... can you take it from there?
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  9. #9
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    Hi,

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

    Thank you
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  10. #10
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    '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?
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  11. #11
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    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
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  12. #12
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    Quote Originally Posted by Krizalid View Post
    \frac{{x^2 - x + 6}}<br />
{{x\left( {x^2 + 3} \right)}} = \frac{2}<br />
{x} - \frac{{x + 1}}<br />
{{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.
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  13. #13
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    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
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    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
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  15. #15
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    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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