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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= - \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 = 1 - x

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

Simplify

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

Add a constant to the solution

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

\int \frac{1}{x ( 1 + x )} d x

\int \frac{x + 1 + tanh ^{2} ( \frac{1}{x} )}{x ^{2}} d x

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

\int \frac{x - 1}{x ^{2} + 2 x + 3} d x

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