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

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Author Comment/Response
Bill Simpson
07/11/13 3:02pm

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]

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Subject (listing for 'Integrate difference btw v8 & v9')
Author Date Posted
Integrate difference btw v8 & v9 Timur 07/11/13 09:54am
Re: Integrate difference btw v8 & v9 Bill Simpson 07/11/13 3:02pm
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