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Math Help - Help with Integrating: xln(1+x)dx using substitution.

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    Help with Integrating: xln(1+x)dx using substitution.

    I am having a difficult time working this Integral. The instructions are to first make a substitution and then integrate by parts. I understand both, but I feel this one has me working in circles. Any insight would be appreciated.

    The indefinite integral (most general) of xln(1+x)dx
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    Re: Help with Integrating: xln(1+x)dx using substitution.

    Make the substitution u = (x+1). This will leave you with:

    (u-1)ln(u). Deal separately with u*ln(u) and -ln(u). Each of them can be integrated using integration by parts. In the first case, integrate u and differentiate ln(u); the product of this will be a linear function easy to integrate. The second one is similar, write it as -1 * ln(u). I'll leave the details to you, tell me if this was too vague.
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    Lightbulb Re: Help with Integrating: xln(1+x)dx using substitution.

    How about: x\to e^u-1
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    Re: Help with Integrating: xln(1+x)dx using substitution.

    Let me show you what I've been trying.

    I don't see how to make an integral sign. So I will use a normal 'S'

    S xln(1+x)dx Let t = 1 + x then x = t - 1
    so dt = (1/2)x2dx and dx = (2/x2)dt

    so in terms of t:

    S (t - 1)ln(t)(2/(t-1))2dt = S [2(t - 1) / (t - 1)2]ln(t)dt = 2S [1 / (t - 1)ln(t)dt

    everything I get after this seems wrong. No matter how I manipulate it. Did I substitute incorrectly?
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    Re: Help with Integrating: xln(1+x)dx using substitution.

    Quote Originally Posted by iAmKrizzle View Post
    Let me show you what I've been trying.

    I don't see how to make an integral sign. So I will use a normal 'S'

    S xln(1+x)dx Let t = 1 + x then x = t - 1
    so dt = (1/2)x2dx and dx = (2/x2)dt

    so in terms of t:

    S (t - 1)ln(t)(2/(t-1))2dt = S [2(t - 1) / (t - 1)2]ln(t)dt = 2S [1 / (t - 1)ln(t)dt

    everything I get after this seems wrong. No matter how I manipulate it. Did I substitute incorrectly?
    Last time I checked, if t = 1 + x, then dt/dx = 1 and so dt = dx...
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    Re: Help with Integrating: xln(1+x)dx using substitution.

    Hello, iAmKrizzle!

    I don't know why they suggested a substitution first.


    I \;=\; \int x\ln(1+x)\,dx

    \text{By parts: }\;\begin{Bmatrix}u &=& \ln(1+x) && dv &=& x\,dx \\ du &=& \dfrac{dx}{1+x} && v &=& \frac{1}{2}x^2 \end{Bmatrix}

    I \;=\;\tfrac{1}{2}x^2\ln(1+x) - \tfrac{1}{2}\int\frac{x^2}{1+x}\,dx

    I \;=\;\tfrac{1}{2}x^2\ln(1+x) - \tfrac{1}{2}\int\left(x - 1 + \frac{1}{1+x}\right)dx

    I \;=\;\tfrac{1}{2}x^2\ln(1+x) - \tfrac{1}{2}\left[\tfrac{1}{2}x^2 - x + \ln(1+x)\right] + C

    I \;=\;\tfrac{1}{2}x^2\ln(1+x) - \tfrac{1}{4}x^2 + \tfrac{1}{2}x - \tfrac{1}{2}\ln(1+x) + C

    I \;=\;\tfrac{1}{2}(x^2-1)\ln(1+x) - \tfrac{1}{4}x^2 + \tfrac{1}{2}x + C
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