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1(32v)2+3432

 Dec 7, 2014

Best Answer 

 #1
avatar+118696 
+5

$1(32v)2+3432$

 

I assume that this is meant to be

 

321(32v)2+34dv=32134v2+34dv=32134(v2+1)dv=3×42×31(v2+1)dv=2311+v2dv=23tan1v+c

 

I think that is correct.  

 Dec 9, 2014
 #1
avatar+118696 
+5
Best Answer

$1(32v)2+3432$

 

I assume that this is meant to be

 

321(32v)2+34dv=32134v2+34dv=32134(v2+1)dv=3×42×31(v2+1)dv=2311+v2dv=23tan1v+c

 

I think that is correct.  

Melody Dec 9, 2014

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