Integration (1 Viewer)

[bladeys]

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]

 

[asianese]

Use a substitution.

 

[asianese]

And go from there.

 

[phoenix159]

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

 

[bladeys]

there's a way to do it without using a substitution: use the fact that a = e^ln(a)
i.e. 2 = e^ln(2)
So 2^ln(x) = e^ln(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

 

[asianese]

Then treat ln2 as n, so what's the integral of x^n?

 

[bladeys]

Then treat ln2 as n, so what's the integral of x^n?

Cheers mate, i appreciate it