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Math Help - Today's calculation of integral #1

  1. #1
    Math Engineering Student
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    Today's calculation of integral #1

    Find \int_0^1x\ln(1+x)\,dx.
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    Find \int_0^1x\ln(1+x)\,dx.
    \int_0^1 x\ln (x+1) dx = \int_0^1 \left[ x\cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n} \right] dx= \sum_{n=0}^{\infty} \left[ \int_0^1 \frac{(-1)^n x^{n+1}}{n} \right]
    Now just simplify.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Krizalid View Post
    Find \int_0^1x\ln(1+x)\,dx.
    by parts u=ln(x+1) dv=xdx

    \frac{1}{2}x^2ln(x+1)|_{0}^{1}-\frac{1}{2}\int_{0}^{1}\frac{x^2}{x+1}dx

    rewriting the integrand we get

    \frac{ln(2)}{2}-\frac{1}{2}\int_{0}^{1}\frac{x^2-1+1}{x+1}dx

    \frac{ln(2)}{2}-\frac{1}{2}\int_{0}^{1}\frac{(x-1)(x+1)+1}{x+1}dx=\frac{ln(2)}{2}-\frac{1}{2}\left[ \int_{0}^{1}(x-1)dx+\int_{0}^{1}\frac{1}{x+1}dx \right]

    \frac{ln(2)}{2}-\frac{1}{2}[\frac{1}{2}x^2-x+ln(x+1)|_{0}^{1}]

    \frac{ln(2)}{2}-\frac{1}{2}[\frac{1}{2}-1+ln(2)]=\frac{1}{4}
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  4. #4
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    I knew Hacker would use series, I was thinkin' the same too, even I like double integration tricks:

    \int_0^1 {x\ln (1 + x)\,dx}  = \int_0^1 \!{\int_0^x {\frac{x}<br />
{{1 + y}}\,dy} \,dx} .

    Now by reversing integration order the rest is routine. (And yes, the answer is \frac14.)
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