What is the integral from 0 to x of 1/((1-x)^3) ?

The integral from 0 to x of 1/((1-x)^3) is 1/(2(1-x)^2)-1/2

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integral from 0 to x of 1/((1-x)^3)

\int_{0}^{x} \frac{1}{( 1 - x ) ^{3}} d x

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

Solution steps

\int_{0}^{x} \frac{1}{( 1 - x ) ^{3}} d x

Apply u-substitution

: \int_{1}^{1 - x} - \frac{1}{u ^{3}} d u

= \int_{1}^{1 - x} - \frac{1}{u ^{3}} d u

Take the constant out:

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

= - \int_{1}^{1 - x} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{1}^{1 - x}

= - [- \frac{1}{2 u ^{2}}]_{1}^{1 - x}

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to x of 1/((1-x)^3) ?
The integral from 0 to x of 1/((1-x)^3) is 1/(2(1-x)^2)-1/2