What is the integral of (4^x-sec^2(x)) ?

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

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integral of (4^x-sec^2(x))

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

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

Solution steps

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

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

\int sec^{2} (x) d x = tan (x)

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

Add a constant to the solution

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

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

x \to 0 lim (\frac{e ^{2 x} - 1}{x})

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}, y (0) = 1

\int \frac{x ^{5}}{x ^{6} - 7} d x

f (x) = arccos (4 x^{2})

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