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Math Help - Substitution THEN Integration By Parts

  1. #1
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    Substitution THEN Integration By Parts

    First I have to use substitution on this, then integration by parts.

    ∫ x^3 ln(x^2 + 1) dx

    My guess is that y = x^2 + 1

    Then dy = 2x dx but then I'm stuck since 2x isn't a 3 degree exponent like in the original integral.
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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    First I have to use substitution on this, then integration by parts.

    ∫ x^3 ln(x^2 + 1) dx

    My guess is that y = x^2 + 1

    Then dy = 2x dx but then I'm stuck since 2x isn't a 3 degree exponent like in the original integral.
    No substitution here, just integration by parts.
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  3. #3
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    First I have to use substitution on this, then integration by parts.

    ∫ x^3 ln(x^2 + 1) dx

    My guess is that y = x^2 + 1

    Then dy = 2x dx but then I'm stuck since 2x isn't a 3 degree exponent like in the original integral.
    Therefore dx = \frac{dy}{2x} and so the integral becomes:

    \int x^3 \ln y \, \frac{dy}{2x} = \frac{1}{2} \int x^2 \ln y \, dy = \frac{1}{2} \int (y + 1) \ln y \, dy.

    Now use parts.
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  4. #4
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    Hello, SportfreundeKeaneKent!

    \int x^3\ln(x^2+1)\,dx
    By parts . . .

    . . \begin{array}{ccccccc}u &=& \ln(x^2+1) & \quad & dv &=& x^3\,dx \\ \\[-3mm] du &=&\dfrac{2x}{x^2+1}\,dx & \quad & v &=&\frac{1}{4}x^4 \end{array}


    We have: . \frac{1}{4}x^4\ln(x^2+1) - \int\left(\frac{1}{4}x^4\right)\left(\frac{2x}{x^2  +1}\,dx\right)

    . . . . . . = \;\frac{1}{4}x^4\ln(x^2+1) - \frac{1}{2}\int\frac{x^5}{x^2+1}\,dx

    . . . . . . = \;\frac{1}{4}x^4\ln(x^2+1) - \frac{1}{2}\left(x^3 - x + \frac{x}{x^2+1}\right)\,dx

    Got it?

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  5. #5
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    That way is simpler Soroban but on a test when it says use sub first then I won't be allowed to just use integration by parts.

    Btw Mr_Fantastic, wouldn't the last line on your answer be 1/2∫(y-1)lny dy. y=x^2 + 1 so x^2 = y-1
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  6. #6
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    That way is simpler Soroban but on a test when it says use sub first then I won't be allowed to just use integration by parts.

    Btw Mr_Fantastic, wouldn't the last line on your answer be 1/2∫(y-1)lny dy. y=x^2 + 1 so x^2 = y-1
    Correct. Thankyou for realising it was a mistake instead of panicking and saying that you're all confused now and just don't know what to do
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