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What is $ \displaystyle\int_0^1 \frac{\ln(1+bx)}{1+x} dx $?

I call it $f(b)$ and differentiate with respect to be $b,$ a bit of partial fractions and the $x$ integral can be done. Then I integrate with respect to $b$ and get a bit lost.

Can some of the terms be expressed in terms of dilogarithms? I get lost in the details! Could we avoid all this just go straight to dilogarithms (with a cunning substitution)?

  • 0
    Check out [this question](http://math.stackexchange.com/questions/117246/evaluation-of-the-integral-int-01-frac-ln1-x1-xdx#comment272398_117246) where $b = -1$.2012-03-08
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    Well, if you substitute $bx = b - bu - 1$, you will get $b \int \frac{\ln (1-u)}{bu + (2b-1)} dx + b \int \frac{\ln b}{bu + (2b-1)}$. Not sure how to integrate the first one though.2012-03-08
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    Can you show your working? Seems like differentiating wrt $b$ might work.2012-03-08
  • 0
    Maple says$-dilog \biggl(\frac{b}{b - 1}\biggr) + dilog \biggl(\frac{2 b}{b - 1}\biggr) + \operatorname{ln} (1 + b) \operatorname{ln} \biggl(\frac{2 b}{b - 1}\biggr)$2012-03-08

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