How to integrate 2^x * e^x

[jaykeegan]

I'm wondering how to integrate something of the form 2^x * e^x.


I tried using integration by parts, ∫u dv = v*u - ∫v du. The problem is no matter which term I choose to integrate and which to differentiate, the final solution still has an integral with a product of terms in the form (2^x) and (e^x). (∫2^x = 2^x / ln2 and (2^x)' = 2^x * ln2).

 

[haruspex]

Rewrite as e^u, where u is a function of x.

 

[Dick]

I'm wondering how to integrate something of the form 2^x * e^x.


I tried using integration by parts, ∫u dv = v*u - ∫v du. The problem is no matter which term I choose to integrate and which to differentiate, the final solution still has an integral with a product of terms in the form (2^x) and (e^x). (∫2^x = 2^x / ln2 and (2^x)' = 2^x * ln2).

$2^x=e^{log(2) x}$.

 

[jaykeegan]

Ah okay thanks people. Both of these methods will simplify it. It's been so long since I've had to do calculus, haha. Forgotten all the different ways of looking at an expression.

 

[PeroK]

I found this quotation from Laplace, which you might like:

"Nature laughs at the difficulties of integration."