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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}}

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

Substitute back

u = x - 1

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

Add a constant to the solution

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

y^{^{'}} + 2 \cdot y = t

\frac{d}{d x} (ln (x + 4 + x^{2}))

\frac{\partial}{\partial x} (e^{x} ln (x y))

2 x^{3} - 3 x^{2} - 12 + 8

n = 1 \sum \infty 7^{n} x^{n - 1}

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