How to Integrate a Complex Exponential Function with Natural Logarithms?

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Homework Statement



integrate:13((4^x)+(3^x))dx

Homework Equations





The Attempt at a Solution



I know the solution is 13((4^x)/ln(4) + (3^x)/ln(3)) + C

Can someone explain to me how this works? I don't know where the ln's are coming from. How would I differentiate this back to the original function?
 
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Are you aware of the exponential function ex which can be defined by the the following property?\frac{d}{dx}e^x = e^xGiven that, consider something like 4x. By the definition of the natural logarithm and its inverse, the exponential function, you can write "4" as eln4. Therefore, our function becomes:4^x = (e^{\ln4})^x = e^{x\ln4}It follows from the chain rule that\frac{d}{dx}(4^x) = \frac{d}{dx}(e^{x\ln4}) = \ln4 e^{x\ln4} = (\ln4)4^xNow, in this example it didn't matter that the number was 4. It could have been anything. So, using the properties of the exponential function and the natural logarithm, we have shown that, for any number "a", it is true that:\frac{d}{dx}a^x = (\ln a) a^x
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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