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

The integral of (1/4)^{x+1} is -1/(ln(2)*2^{2x+3)}+C

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

\int (\frac{1}{4})^{x + 1} d x

- \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 3}} + C

Solution steps

\int (\frac{1}{4})^{x + 1} d x

(\frac{1}{4})^{x + 1} = (\frac{1}{4})^{1} (\frac{1}{4})^{x}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= (\frac{1}{4})^{1} \cdot \int (\frac{1}{4})^{x} d x

Simplify

= \frac{1}{4} \cdot \int (\frac{1}{4})^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{4} \cdot \frac{( \frac{1}{4} ) ^{x}}{ln ( \frac{1}{4} )}

Simplify

\frac{1}{4} \cdot \frac{( \frac{1}{4} ) ^{x}}{ln ( \frac{1}{4} )} : - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 3}}

= - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 3}}

Add a constant to the solution

= - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 3}} + C

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

\int (\frac{x}{1 + x ^{2}})^{2} d x

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

x \to - 1 + lim (\frac{x}{x + 1})

\frac{\partial}{\partial t} (3 t)

What is the integral of (1/4)^{x+1} ?
The integral of (1/4)^{x+1} is -1/(ln(2)*2^{2x+3)}+C