Hard Problem 79D

$\dfrac{\text{d}}{\text{d}x} n^{x} = n^{x} ln(n)$

Then by substitution $\displaystyle\int ln(4) 4^{x} dx =\displaystyle\int (\dfrac{\text{d}}{\text{d}x} 4^{x}) dx = \displaystyle\int 1 \times d(4^{x}) = 4^{x} + C$
From the rule that $\displaystyle\int 1 du = u +C$

Hence $\displaystyle\int_{2}^{3} ln(4) 4^{x} = (4^{3} +C -(4^{2} +C)) = 4^{3} - 4^{2} = 48$

$\int 4^x dx=\frac{4^x}{\ln(4)}+C_1$ $\int 4^x\ln(4)dx=4^x+C_2$ $\int_{2}^{3}4^x\ln(4)dx=4^3-4^2=48$