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Student Support Forum > General > > "Integrate difference btw v8 & v9"

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Please answer this:2+5 =



Original Message (ID '552287') By Bill Simpson:
FunctionExpand, and perhaps FullSimplify, can do this, but only when they have "simple" exact integers for the first three arguments of Hypergeometric2F1, so use a little substitution. VERSION 8 Integrate[((22.20449885*(0.002343867)*p^(1/0.789890201))/(1 + (0.002343867)*p^(1/0.789890201)))/p, p] 17.5391 Log[1. + 0.00234387 p^1.266] Integrate[((0.602587148*(8.304501705)*p^(1/0.853929364))/(1 + (8.304501705)*p^(1/0.853929364)) + (3.040496738*0.505954709*p^(1/0.889250767))/(1 + (0.505954709*p^(1/0.889250767))))/p, p] 2.70376 Log[1. + 0.505955 p^1.12454] + 0.514567 Log[1. + 8.3045 p^1.17106] VERSION 9 In[1]:= FunctionExpand[Integrate[((22.20449885*(0.002343867)*p^(1/0.789890201))/(1 + (0.002343867)*p^(1/0.789890201)))/p, p] /. { 1. -> 1, 2. -> 2}] Out[1]= 17.5391 Log[1 + 0.00234387 p^1.266] In[2]:= FunctionExpand[Integrate[((0.602587148*(8.304501705)*p^(1/0.853929364))/(1 + (8.304501705)*p^(1/0.853929364)) + (3.040496738*0.505954709*p^(1/0.889250767))/(1 + (0.505954709*p^(1/0.889250767))))/p, p] /. { 1. -> 1, 2. -> 2}] Out[2]= 2.70376 Log[1 + 0.505955 p^1.12454] + 0.514567 Log[1 + 8.3045 p^1.17106]