Integration (1 Viewer)

bladeys

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How do you integrate 2^(logx)?

Ive got it down to [x*2^(logx)]/log(2) but wolfram alpha says the answer is x*2^(logx)]/[log(2)+1]
 

phoenix159

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there's a way to do it without using a substitution: use the fact that a = eln(a)
i.e. 2 = eln(2)
So 2ln(x) = eln(x)*ln(2)
then integrate the RHS to get the answer
 

bladeys

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there's a way to do it without using a substitution: use the fact that a = eln(a)
i.e. 2 = eln(2)
So 2ln(x) = eln(x)*ln(2)
then integrate the RHS to get the answer
Yep this is what i did but I dont think i integrated it properly because I got [x*2^(logx)]/log(2) instead of x*2^(logx)]/[log(2)+1]
Im not sure where the 1 on the denominator comes from
 

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