Integral

**jones123** · December 29th 2012, 10:05 AM

Hello! How would you integrate 2^x / ln2? Thanks already!

**a tutor** · December 29th 2012, 10:21 AM

$\displaystyle 2^x=e^{\ln 2^x}=e^{x \ln 2}$

**Deveno** · December 29th 2012, 10:38 AM

note that:

$\frac{1}{\ln(2)} = \frac{\ln(2)}{(\ln(2))^2}$

you can take the denominator (which is just a constant) outside the integral, so:

$\int \frac{2^x}{\ln(2)}\ dx = \frac{1}{(\ln(2))^2}\int \ln(2)2^x\ dx = \frac{1}{(\ln(2))^2} \int e^{\ln(2)x} \ln(2)\ dx$

if you use the substitution $u = \ln(2)x$ what would $du$ be?

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