From: drhuang57@gmail.com   
      
   On Thursday, 8 June 2017 19:31:51 UTC+10, asetof...@gmail.com  wrote:   
   > Int(f(2*x+1)dx)==Int((f(2*x+1)/2)*2dx)   
   > ==Int(f(2*x+1)/2 d(2*x+1))==   
   > ==Subst(Int(f(y)/2 dy),y=2*x+1)   
   > ==F(2*x+1)/2 + c   
   > Where F'(x)=f(x)   
   > It is easy only note that can be written using y==2*x+ 1 as variable... Or   
   2*x+1 is just a costant...   
      
   You need to define int(f(2x+1),x), int(f(2x+2),x), int(2f(2x+3),x),   
   int(3f(3x+1),x), int(4f(3x+2),x) ... etc.? How many rules did you need to   
   define? You need to define only one rule in mathHandbook, it auto learn these   
   all.   
      
   It learns the many complicated integrals, e.g. 2f(2x+1), from a single simple   
   integral of f(x).   
      
   reference math handbook   
   www.mathHandbook.com   
      
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