How to find the integral $\int4^{-x}dx$?

Question

What approach would be ideal in finding the integral $\int4^{-x}dx$?

Answer 1

Firstly, the most Important thing here is that $(4^{-x})'=(-\ln 4)4^{-x}$

So we rewrite our integral as follows:

$\int 4^{-x}\,dx=\frac{-1}{\ln 4}\int -\ln 4\cdot 4^{-x}\,dx=\frac{-1}{\ln 4}\int (4^{-x})'\,dx=-\frac{1}{\ln 4}4^{-x}+C$.

Answer 2

Rewrite the integral as $$\int e^{(- \ln 4) x} \mathrm{d}x. $$

Answer 3

Hint: let $u = -x$, then you can go from here.

Answer 4

$$ \int 4^{-x}\,{\rm d}x = -\,{1 \over \ln(4)}\int\left[-\ln(4)\,4^{-x}\right]\,{\rm d}x =-\,{1 \over \ln(4)}\int{{\rm d 4^{-x}} \over {\rm d}x}\,{\rm d}x =-\,{4^{-x} \over \ln(4)} + \mbox{a constant} $$