What is the integral of 1/((1-x)^n) ?

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

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integral of 1/((1-x)^n)

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

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

Solution steps

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

Apply exponent rule:

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

\frac{1}{( 1 - x ) ^{n}} = (1 - x)^{- n}

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

Apply u-substitution

: \int - u^{- n} d u

= \int - u^{- n} d u

Take the constant out:

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

= - \int u^{- n} d u

Apply the Power Rule

: \frac{u ^{- n + 1}}{- n + 1}

= - \frac{u ^{- n + 1}}{- n + 1}

Substitute back

u = 1 - x

= - \frac{( 1 - x ) ^{- n + 1}}{- n + 1}

Add a constant to the solution

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

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