What is the integral of p(1-p)^{x-1} ?

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

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integral of p(1-p)^{x-1}

\int p (1 - p)^{x - 1} d x

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

Solution steps

\int p (1 - p)^{x - 1} d x

Take the constant out:

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

= p \cdot \int (1 - p)^{- 1 + x} d x

(1 - p)^{- 1 + x} = (1 - p)^{- 1} (1 - p)^{x}

= p \cdot \int (1 - p)^{- 1} (1 - p)^{x} d x

Take the constant out:

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

= p (1 - p)^{- 1} \cdot \int (1 - p)^{x} d x

Apply exponent rule:

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

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

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

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

Simplify

p \frac{1}{1 - p} \cdot \frac{( 1 - p ) ^{x}}{ln ( 1 - p )} : \frac{p ( 1 - p ) ^{x - 1}}{ln ( 1 - p )}

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

Add a constant to the solution

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

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