What is the integral of 1/((1-x)^6) ?

The integral of 1/((1-x)^6) is -1/(5(x-1)^5)+C

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integral of 1/((1-x)^6)

\int \frac{1}{( 1 - x ) ^{6}} d x

- \frac{1}{5 ( x - 1 ) ^{5}} + C

Solution steps

\int \frac{1}{( 1 - x ) ^{6}} d x

Take the partial fraction of

\frac{1}{( 1 - x ) ^{6}} : \frac{1}{( x - 1 ) ^{6}}

= \int \frac{1}{( x - 1 ) ^{6}} d x

Apply u-substitution

: \int \frac{1}{u ^{6}} d u

= \int \frac{1}{u ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= - \frac{1}{5 u ^{5}}

Substitute back

u = x - 1

= - \frac{1}{5 ( x - 1 ) ^{5}}

Add a constant to the solution

= - \frac{1}{5 ( x - 1 ) ^{5}} + C

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\frac{- 8 x}{x ^{2} + 1} (0)

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What is the integral of 1/((1-x)^6) ?
The integral of 1/((1-x)^6) is -1/(5(x-1)^5)+C