What is the integral of 1/(4^{2x)} ?

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

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integral of 1/(4^{2x)}

\int \frac{1}{4 ^{2 x}} d x

16 x + C

Solution steps

\int \frac{1}{4 ^{2 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{4 ^{2 x}} = 4^{- 2 x}

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

Apply u-substitution

: \int - 8 d u

= \int - 8 d u

Integral of a constant:

\int a d x = a x

= (- 8) u

Substitute back

u = - 2 x

= (- 8) (- 2 x)

Simplify

= 16 x

Add a constant to the solution

= 16 x + C

ln (ln (4 s))

\frac{\partial}{\partial x} (x^{3} + 2 x y + y^{2})

\frac{d}{d x} (0.015 + x)

f (x) = 6 x^{2} - 10 x + 4, at x = 1

x \to 3 lim (ln (x))