What is the integral of x/((1-x)^4) ?

The integral of x/((1-x)^4) is -1/(2(x-1)^2)-1/(3(x-1)^3)+C

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integral of x/((1-x)^4)

\int \frac{x}{( 1 - x ) ^{4}} d x

- \frac{1}{2 ( x - 1 ) ^{2}} - \frac{1}{3 ( x - 1 ) ^{3}} + C

Solution steps

\int \frac{x}{( 1 - x ) ^{4}} d x

Take the partial fraction of

\frac{x}{( 1 - x ) ^{4}} : \frac{1}{( x - 1 ) ^{3}} + \frac{1}{( x - 1 ) ^{4}}

= \int \frac{1}{( x - 1 ) ^{3}} + \frac{1}{( x - 1 ) ^{4}} d x

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}{( x - 1 ) ^{3}} d x + \int \frac{1}{( x - 1 ) ^{4}} d x

\int \frac{1}{( x - 1 ) ^{3}} d x = - \frac{1}{2 ( x - 1 ) ^{2}}

\int \frac{1}{( x - 1 ) ^{4}} d x = - \frac{1}{3 ( x - 1 ) ^{3}}

= - \frac{1}{2 ( x - 1 ) ^{2}} - \frac{1}{3 ( x - 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{2 ( x - 1 ) ^{2}} - \frac{1}{3 ( x - 1 ) ^{3}} + C

x \to 3 lim (x^{2} - 2 x)

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

y = \frac{t}{( t - 1 ) ^{2}}

f (x) = \frac{4}{5 x + 1}, at x = - 1

What is the integral of x/((1-x)^4) ?
The integral of x/((1-x)^4) is -1/(2(x-1)^2)-1/(3(x-1)^3)+C