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Math Help - Help with simple (i think...) integral

  1. #1
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    Talking Help with simple (i think...) integral

    Integral of.......

    3(u+1)
    u^2 +1 du

    it started as the integral of...

    3x
    x^2-2x+2

    and i completed the square to get the integral of...

    3x
    (x-1)^2 +1

    Then put u = x-1, so that x = u+1, and thats how i got to where i am now... i always get stuck at this part...thanks
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  2. #2
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    \int \frac{3(u+1)}{u^2+1} \, du = \int \frac{3u}{u^2+1} \, du \, +  \, \int \frac{3}{u^2+1} \, du = \frac{3}{2}\int \frac{2u}{u^2+1} \, du \, + \, 3\int \frac{1}{u^2+1} \, du

    can you integrate the sum of the last two integral expressions?
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  3. #3
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    Hello, Smac!

    It started as: . \int\frac{3x}{x^2-2x+2}\,dx

    and i completed the square to get: . \int \frac{3x}{(x-1)^2+1}\,dx . . . . Good!

    Then let u \:=\: x-1 ... and got: . \int\frac{3(u+1)}{u^2+1}\,du . . . . Right!
    Make two integrals . . .

    . . 3\left[\int \frac{u+1}{u^2+1}\,du\right] \;=\;3\left[\underbrace{\int \frac{u\,du}{u^2+1}}_{\ln}   + \underbrace{\int\frac{du}{u^2+1}}_{\arctan}\,\righ  t]

    Got it?

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  4. #4
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    its like my brain freezes when i get an integral as such.

    The integral of...

    (3) [ (u du) / (u^2 +1) ]

    you said it was the natural log, but if you do the natural log, dont you need a constant as the numerator?

    the way i did it ended up using r = u^2+1 and the integral turns to...

    (3) [ (u dr) / r (2u) ]

    then the "u" cancels to get the integral of...

    (3/2) [ dr / r ]

    which equals.....

    3/2 ln abs(r)
    =>
    3/2 ln abs(u^2 +1)
    =>
    3/2 ln abs( (x+1)^2 + 1)
    =>
    3/2 ln abs(x^2 -2x +1)....which the function in the ln was the denominator all along...to me this seems wrong but the math seems correct...i probably missed something...(besides the constant at the end of course)
    ???
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  5. #5
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    the integral that yields a a natural log is \int \frac{f'(x)}{f(x)} dx

    I set it up for you in my previous post ...

    \frac{3}{2} \int  \frac{2u}{u^2+1}  \, du  =  \frac{3}{2}  \ln(u^2 + 1)  + C
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  6. #6
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    yeah i got it, but i just did so many "u-subs" that i get confused with my work. Now that you set it up as...

    Int of - - f '(x) / f(x) = ln abs(f(x)) + c makes it easier, thanks
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