What is the integral of (cos(x)+4^x) ?

The integral of (cos(x)+4^x) is sin(x)+(2^{2x-1})/(ln(2))+C

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integral of (cos(x)+4^x)

\int (cos (x) + 4^{x}) d x

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

Solution steps

\int cos (x) + 4^{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 cos (x) d x + \int 4^{x} d x

\int cos (x) d x = sin (x)

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

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

Add a constant to the solution

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

\int \frac{x ^{- 1}}{- 1} d x

\int \frac{( a ^{x} ln ( a ) )}{( a ^{x} + 1 )} d x

\int \frac{8}{( x ^{2} - 4 )} d x

\int \frac{( e ^{x} e ^{x} - 1 )}{e ^{x} + 8} d x

\int \frac{1}{( y )} d y

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