What is the integral of (10)/((x-1)^2) ?

The integral of (10)/((x-1)^2) is -(10)/(x-1)+C

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integral of (10)/((x-1)^2)

\int \frac{10}{( x - 1 ) ^{2}} d x

- \frac{10}{x - 1} + C

Solution steps

\int \frac{10}{( x - 1 ) ^{2}} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{( x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= 10 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 10 (- \frac{1}{u})

Substitute back

u = x - 1

= 10 (- \frac{1}{x - 1})

Simplify

10 (- \frac{1}{x - 1}) : - \frac{10}{x - 1}

= - \frac{10}{x - 1}

Add a constant to the solution

= - \frac{10}{x - 1} + C

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What is the integral of (10)/((x-1)^2) ?
The integral of (10)/((x-1)^2) is -(10)/(x-1)+C