Is the Integral of ln(1+2^x) Correctly Calculated?

  • Thread starter UrbanXrisis
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In summary, the integral of ln(1+2^x) can be solved by using substitution and integration by parts. However, the result obtained by differentiating the original function appears to be the same as the integral, which means there may be a mistake in the process.
  • #1
UrbanXrisis
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what is the integral of [tex]ln(1+2^x)[/tex]

[tex]\int ln(1+2^x)=\frac{2^xln(2)}{1+2^x} [/tex]

is this correct?
 
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  • #2
Differentiate and see what you get.
 
  • #3
Try

[tex] u = 2^x+1, du = 2^xln(2) [/tex]

Then

[tex] \int ln(1+2^x) dx = ln(2)\int (u-1)ln(u) du [/tex]

Which can be done by parts.
 
  • #4
UrbanXrisis said:
what is the integral of [tex]ln(1+2^x)[/tex]

[tex]\int ln(1+2^x)=\frac{2^xln(2)}{1+2^x} [/tex]

is this correct?

That result you got looks suspiciously like the derivative of [itex] \ln(1+2^x) [/itex]! What I'm saying is, it looks like you differentiated using the chain rule:

let u = 1 + 2^x

[tex] \frac{d}{dx}[\ln(1+2^x)] = \frac{d}{dx}(\ln u) [/tex]

[tex] = \frac{1}{u} \frac{du}{dx} [/tex]

[tex] = \left(\frac{1}{1+2^x}\right) \frac{d}{dx}(1+2^x) [/tex]

[tex] = \frac{(\ln2)2^x}{1+2^x} [/tex]

But you were supposed to integrate! :smile:

I just thought I'd point that out, so you could see the mistake.
 

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