What is the integral of (16^x+20^x)/(4^x) ?

The integral of (16^x+20^x)/(4^x) is (2^{2x-1})/(ln(2))+(5^x)/(ln(5))+C

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integral of (16^x+20^x)/(4^x)

\int \frac{1 6 ^{x} + 2 0 ^{x}}{4 ^{x}} d x

\frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{5 ^{x}}{ln ( 5 )} + C

Solution steps

\int \frac{1 6 ^{x} + 2 0 ^{x}}{4 ^{x}} d x

Expand

\frac{1 6 ^{x} + 2 0 ^{x}}{4 ^{x}} : 4^{x} + 5^{x}

= \int 4^{x} + 5^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4^{x} d x + \int 5^{x} d x

\int 4^{x} d x = \frac{2 ^{2 x - 1}}{ln ( 2 )}

\int 5^{x} d x = \frac{5 ^{x}}{ln ( 5 )}

= \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{5 ^{x}}{ln ( 5 )}

Add a constant to the solution

= \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{5 ^{x}}{ln ( 5 )} + C

x y^{^{'}} + 3 y - 3 = 0

\int \frac{ln ( x )}{x} - \frac{( ln ( x ) ) ^{2}}{x} d x

\int_{0}^{π} (4 e^{x} + 7 sin (x)) d x

\int \frac{23}{x ^{3} - 27} d x

\frac{d}{d x} (arctan (\frac{2 x}{1 - x ^{2}}))

What is the integral of (16^x+20^x)/(4^x) ?
The integral of (16^x+20^x)/(4^x) is (2^{2x-1})/(ln(2))+(5^x)/(ln(5))+C