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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x}{( 1 - x ) ^{4}} - \int \frac{4 x}{( 1 - x ) ^{5}} d x

= \frac{x}{( 1 - x ) ^{4}} - \int \frac{4 x}{( 1 - x ) ^{5}} d x

\int \frac{4 x}{( 1 - x ) ^{5}} d x = - \frac{4}{3 ( 1 - x ) ^{3}} + \frac{1}{( 1 - x ) ^{4}}

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (8 e^{y x} - x^{2} y^{3})

\int 10 e^{- 10 x} d x

\int - \frac{8}{π} x d x

f (x) = \frac{2}{1 + e ^{- x}}, (0, 1)

x^{^{'}} = (x - 17)^{2}

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