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Math Help - An easy indefinite integral

  1. #1
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    An easy indefinite integral

    Solve \int {\frac{{dx}}<br />
{{\sqrt {ax^2  + bx + c} }}} .\quad b,c \in \mathbb{R},\,a > 0.
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    Quote Originally Posted by Krizalid View Post
    Solve \int {\frac{{dx}}<br />
{{\sqrt {ax^2  + bx + c} }}} .\quad b,c \in \mathbb{R},\,a > 0.
    It is really easy.

    You can write ax^2+bx+c = a\left( x^2 + \frac{b}{a}+\frac{c}{a}\right) = a\left[ \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a} \right]

    Now it depends whether 4ac -b^2 is positive, negative, or zero.
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    Well, that's not actually the solution what I was lookin' for.
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    Too easy indeed.

    Quote Originally Posted by Krizalid View Post
    Well, that's not actually the solution what I was lookin' for.
    WLOG I'll consider \displaystyle \int \frac{1}{\sqrt{x^2 + Bx + C}} \, dx.

    Then I reckon the solution what you was lookin' for uses the substitution \sqrt{x^2 + Bx + C} = x + z.

    \ln (B + 2x + 2\sqrt{x^2 + Bx + C}) falls out easy peasy.
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  5. #5
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    Well, the substitution what I was lookin' for is u = \frac{{2ax + b}}<br />
{{\sqrt a }} + 2\sqrt {ax^2  + bx + c} .

    The rest follows easily.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    Well, the substitution what I was lookin' for is u = \frac{{2ax + b}}<br />
{{\sqrt a }} + 2\sqrt {ax^2  + bx + c} .

    The rest follows easily.
    Hmmm yes quite obvious, quite obvious *cough*
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    Quote Originally Posted by Krizalid View Post
    Well, the substitution what I was lookin' for is u = \frac{{2ax + b}}<br />
{{\sqrt a }} + 2\sqrt {ax^2  + bx + c} .

    The rest follows easily.
    Quote Originally Posted by DivideBy0 View Post
    Hmmm yes quite obvious, quite obvious *cough*
    It is quite obvious if you look at what I did. Complete the square and substitute the stuff inside the square.
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    Quote Originally Posted by Krizalid View Post
    Well, the substitution what I was lookin' for is u = \frac{{2ax + b}}<br />
{{\sqrt a }} + 2\sqrt {ax^2  + bx + c} .

    The rest follows easily.
    Well, since a > 0 is given, take a factor of a out of the radical and you get what I got (within a factor) - WLOG. Why hoe the hard road (but you do have to solve for x in terms of z with the Euler substitution - not too hard).
    Last edited by mr fantastic; December 25th 2007 at 01:19 PM.
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  9. #9
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    Quote Originally Posted by Krizalid View Post
    Well, the substitution what I was lookin' for is u = \frac{{2ax + b}}<br />
{{\sqrt a }} + 2\sqrt {ax^2  + bx + c} .

    The rest follows easily.
    Following from my previous post (and probably totally not necessary for Krizalid):

    WLOG let a = 1, b = B and c = C. You get \displaystyle u = 2x + B + 2\sqrt {x^2  + Bx + C}. The substitution I used (the standard Euler substitution, just to set the record straight) is \displaystyle u + x = \sqrt{x^2 + Bx + C}. Arguably simpler .....?? Arguably easier ....?? But ...... intuitively way less obvious .. ..??

    Personally, I prefer \displaystyle u + x = \sqrt {x^2  + Bx + C} ..... easy to remember for this type of integrand viz. f(\sqrt{x^2 + Bx + C}) and the algebra for getting the stuff that's needed (dx and x in terms of u) is easy.

    Eg. To solve for x in terms of u: square both sides of \displaystyle \sqrt{x^2 + Bx + C} = x + u, subtract x^2 from both sides, group x-terms, solve for x.


    Note: In my original post I used z (not u) as my new variable, but u's OK I guess.
    Last edited by mr fantastic; December 25th 2007 at 05:47 PM.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    It is quite obvious if you look at what I did. Complete the square and substitute the stuff inside the square.
    It is obvious what you did TPH. It is always nice to see everyone show their own solutions to problems here at MHF!
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