What is the antiderivative of $\frac{x}{{(1 - x)}^{4}}$ $x/(1-x)^4$ ?

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What is the antiderivative of $\frac{x}{{(1 - x)}^{4}}$ $x/(1-x)^4$ ?

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Answer 1 · 2016-06-11T08:37:37

$\int \frac{x}{{(1 - x)}^{4}} d x = \frac{3 x - 1}{6 {(1 - x)}^{3}} + C$ $int x/(1-x)^4 dx = (3x-1)/(6(1-x)^3) + C$

Explanation:

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

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

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

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

So:

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

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

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

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

$= \frac{2}{6 {(1 - x)}^{3}} - \frac{3 (1 - x)}{6 {(1 - x)}^{3}} + C$ $= 2/(6(1-x)^3)-(3(1-x))/(6(1-x)^3) + C$

$= \frac{3 x - 1}{6 {(1 - x)}^{3}} + C$ $= (3x-1)/(6(1-x)^3) + C$

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