What is the integral of 1/((x-1)^5) ?

The integral of 1/((x-1)^5) is -1/(4(x-1)^4)+C

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integral of 1/((x-1)^5)

\int \frac{1}{( x - 1 ) ^{5}} d x

- \frac{1}{4 ( x - 1 ) ^{4}} + C

Solution steps

\int \frac{1}{( x - 1 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{u ^{5}} d u

= \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= - \frac{1}{4 u ^{4}}

Substitute back

u = x - 1

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

Add a constant to the solution

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

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What is the integral of 1/((x-1)^5) ?
The integral of 1/((x-1)^5) is -1/(4(x-1)^4)+C