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Math Help - Integration by substitution

  1. #1
    Senior Member slevvio's Avatar
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    Integration by substitution

    Hello, I have to integrate the following by substitution, but I am having trouble... any help would be appreciated...

     \int x \log (x^2 + 1)dx . thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slevvio View Post
    Hello, I have to integrate the following by substitution, but I am having trouble... any help would be appreciated...

     \int x \log (x^2 + 1)dx . thanks
    i suppose by log here, you mean ln. just make the substitution u = ln(x^2 + 1)
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  3. #3
    Senior Member slevvio's Avatar
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    Thanks for the response however using that substitution I end up with

    \frac{1}{2}\int \ln(u) du , which I don't know how to integrate. Thanks
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    Quote Originally Posted by slevvio View Post
    Thanks for the response however using that substitution I end up with

    \frac{1}{2}\int \ln(u) du , which I don't know how to integrate. Thanks
    That's what happens if you use the substitution: u=x^2 + 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slevvio View Post
    Thanks for the response however using that substitution I end up with

    \frac{1}{2}\int \ln(u) du , which I don't know how to integrate. Thanks
    see here

    i'm not sure that's the integral we would get though...let me see. again, i assume you mean ln when you type log
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    That's what happens if you use the substitution: u=x^2 + 1
    correct, which happens to be easier than my substitution

    well, not really, you'd end up with \frac 12 \int ue^u~du for mine, which is almost the same thing, in terms of difficultly
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  7. #7
    Senior Member slevvio's Avatar
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    hehe i knew I was doing somethign right... thanks a lot, our maths lecturer at uni is German and she uses Log when she means ln in the notes i just copied it straight out
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  8. #8
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    Hello, slevvio!

    I have to integrate the following by substitution. .?

    . . .  \int x \log (x^2 + 1)dx
    This cannot be integrated by substitution alone.

    We must integrate by parts . . .

    . . \begin{array}{ccccccc}u & = & \ln(x^2+1) & \quad & dv & = & x\,dx \\ du & = & \frac{2x}{x^2+1}\,dx & \quad & v & = & \frac{1}{2}x^2 \end{array}

    Then we have: . \frac{1}{2}x^2\ln(x^2+1) - 2\int\frac{x^3}{x^2+1}\,dx

    . . .  =\;\frac{1}{2}x^2\ln(x^2+1) - 2\int\left(x - \frac{x}{x^2+1}\right)\,dx . . . . etc.

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