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

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

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integral of (2x-4^x)

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

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

Solution steps

\int 2 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 2 x d x - \int 4^{x} d x

\int 2 x d x = x^{2}

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

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

Add a constant to the solution

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

y = x^{2} - 4, y = 5

x \to \infty lim (1^{- x})

\frac{d}{d x} (8^{s i n (π x)})

\int_{1}^{8} \frac{2}{x} d x

f (x) = - 6

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