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integral of 4^3x dx
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\[\text{The integraL of }4^{3 x}\text{ is }\frac{4^{3 x}}{3 \ln (4)}\text{}\] \[\frac{4^{3 x}}{3 \ln (4)}+c\]
WOuld you by any chance explain it to me to understand better?
Let u=3x dx=(du)/(3) then, integrate.. (4^(u))/(3)/(ln(4))+C ((64)^(x))/(3)+C
\[\frac{d}{dx} b^x=b^x\ln(b)\] work backwards
e^y = 4 or y = ln 4 4^3x = e^(ln4)3x
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\[\int b^xdx=\frac{b^x}{\ln(b)}\]
i guess it would be pretty much the same.
yeah, and (64)^(x)/(3lnx)+C
thank you
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