ace21a92   
   From: Albert_Rich@msn.com   
      
   An antiderivative of arctanh(t)*log(t)/(t-1) wrt t is:   
      
   -1/2*Log[2]*Log[t]*Log[(1 + t)/2] + 1/4*Log[2]*Log[(1 + t)/2]^2 +   
    1/2*Log[1 - t]*Log[t]*Log[1 + t] -   
    1/2*Log[2]*Log[(1 + t)/2]*Log[(1 + t)/(2*t)] +   
    1/4*Log[1 - t]*Log[(1 + t)/(2*t)]^2 +   
    1/4*Log[-(1/t)]*Log[(1 + t)/(2*t)]^2 -   
    1/4*Log[(-1 + t)/(2*t)]*Log[(1 + t)/(2*t)]^2 +   
    1/2*Log[(1 + t)/(2*t)]*PolyLog[2, 1 + 1/t] -   
    ArcTanh[t]*PolyLog[2, 1 - t] +   
    1/2*Log[1 + t]*PolyLog[2, 1 - t] + 1/2*Log[1 + t]*PolyLog[2, t] -   
    1/2*Log[(1 + t)/(2*t)]*PolyLog[2, t] +   
    1/2*Log[t]*PolyLog[2, (1 + t)/2] +   
    1/2*Log[(1 + t)/(2*t)]*PolyLog[2, (1 + t)/2] -   
    1/2*Log[(1 + t)/(2*t)]*PolyLog[2, (1 + t)/(2*t)] -   
    1/2*PolyLog[3, 1 + 1/t] -   
    1/2*PolyLog[3, 1 - t] - (1/2)*PolyLog[3, t] -   
    1/2*PolyLog[3, (1 + t)/2] +   
    1/2*PolyLog[3, (1 + t)/(2*t)]   
      
   Taking the difference of evaluating its limits as t approaches 1 and 0 gives   
   1/8 pi^2 log(2).   
      
   Albert   
      
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