How do you find the antiderivative of $\frac{1}{{(1 - x)}^{2}}$ $1/(1-x)^2$ ?

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How do you find the antiderivative of $\frac{1}{{(1 - x)}^{2}}$ $1/(1-x)^2$ ?

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Answer 1 · 2015-06-30T20:13:06

I found: $\frac{1}{1 - x} + c$ $1/(1-x)+c$

Explanation:

I tried this:
$\int \frac{1}{{(1 - x)}^{2}} d x =$ $int1/(1-x)^2dx=$
Where: $d [1 - x] = - d x$ $d[1-x]=-dx$ giving:
$\int {(1 - x)}^{- 2} \cdot - d x = \frac{1}{1 - x} + c$ $int(1-x)^-2*-dx=1/(1-x)+c$

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How do you find the antiderivative of $\frac{1}{{(1 - x)}^{2}}$ $1/(1-x)^2$ ?

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