Integrate $3^x$ using the $px + q$ rule

Question

I'm having trouble integrating $3^x$ using the $px + q$ rule. Can some please walk me through this?

Thanks

Answer 1

We can simply take as primitive $$\int a^x \,\text{d}x = \dfrac{a^x}{\log a} + C$$

Suggestion: Verify for yourself that $$\frac{\text{d}}{dx}\left(\frac {a^x}{\log a} + C\right) = a^x$$

So $$\int 3^x \,\text{d}x = \dfrac{3^x}{\log 3} + C.$$

Answer 2

And the $px+q$ rule is...?

You can use $3^x = e^{x\log 3}$ and the obvious change of variable.

Answer 3

If I'm correct in what you think the $px+q$ rule is then take $f(x)=e^x$, $p=\log 3$, and $q=0$, with $\log$ the natural logarithm (of course!).