What is the integral of (8x)/((x-1)^2) ?

The integral of (8x)/((x-1)^2) is 8(ln|x-1|-1/(x-1))+C

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integral of (8x)/((x-1)^2)

\int \frac{8 x}{( x - 1 ) ^{2}} d x

8 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + C

Solution steps

\int \frac{8 x}{( x - 1 ) ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{( x - 1 ) ^{2}} d x

Apply u-substitution

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

= 8 \cdot \int \frac{u + 1}{u ^{2}} d u

Expand

\frac{u + 1}{u ^{2}} : \frac{1}{u} + \frac{1}{u ^{2}}

= 8 \cdot \int \frac{1}{u} + \frac{1}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int \frac{1}{u} d u + \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= 8 (ln ∣ u ∣ - \frac{1}{u})

Substitute back

u = x - 1

= 8 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

Add a constant to the solution

= 8 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + C

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What is the integral of (8x)/((x-1)^2) ?
The integral of (8x)/((x-1)^2) is 8(ln|x-1|-1/(x-1))+C