W WEMG Member Joined Aug 15, 2009 Messages 118 Gender Undisclosed HSC 2011 Dec 21, 2010 #1 Integrate 2^logx. I tried making it base e then integrating but it doesnt work......
D deterministic Member Joined Jul 23, 2010 Messages 423 Gender Male HSC 2009 Dec 21, 2010 #2 2^(logx)= (e^log2)^logx=(e^logx)^log2=x^log2 by use of indices rule in the third step then integrate is easy--> x^(log2+1)/(log2+1) + C EDITTED: forgot the +1 in denominator Last edited: Dec 21, 2010
2^(logx)= (e^log2)^logx=(e^logx)^log2=x^log2 by use of indices rule in the third step then integrate is easy--> x^(log2+1)/(log2+1) + C EDITTED: forgot the +1 in denominator
D Drongoski Well-Known Member Joined Feb 22, 2009 Messages 4,252 Gender Male HSC N/A Dec 21, 2010 #3 Beautiful work! But shouldn't answer be: x^(ln2 + 1)/(ln2 +1) + C ??
W WEMG Member Joined Aug 15, 2009 Messages 118 Gender Undisclosed HSC 2011 Dec 21, 2010 #4 Wow thanks I got it, and I think deterninistic forgot to put the +1 in denominator.
