What is the integral of (5x)/((x-1)^3) ?

The integral of (5x)/((x-1)^3) is 5(-1/(x-1)-1/(2(x-1)^2))+C

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integral of (5x)/((x-1)^3)

\int \frac{5 x}{( x - 1 ) ^{3}} d x

5 (- \frac{1}{x - 1} - \frac{1}{2 ( x - 1 ) ^{2}}) + C

Solution steps

\int \frac{5 x}{( x - 1 ) ^{3}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{( x - 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{u + 1}{u ^{3}} d u

= 5 \cdot \int \frac{u + 1}{u ^{3}} d u

Expand

\frac{u + 1}{u ^{3}} : \frac{1}{u ^{2}} + \frac{1}{u ^{3}}

= 5 \cdot \int \frac{1}{u ^{2}} + \frac{1}{u ^{3}} 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

= 5 (\int \frac{1}{u ^{2}} d u + \int \frac{1}{u ^{3}} d u)

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

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= 5 (- \frac{1}{u} - \frac{1}{2 u ^{2}})

Substitute back

u = x - 1

= 5 (- \frac{1}{x - 1} - \frac{1}{2 ( x - 1 ) ^{2}})

Add a constant to the solution

= 5 (- \frac{1}{x - 1} - \frac{1}{2 ( x - 1 ) ^{2}}) + C

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What is the integral of (5x)/((x-1)^3) ?
The integral of (5x)/((x-1)^3) is 5(-1/(x-1)-1/(2(x-1)^2))+C