What is the integral from 0.43 to x of 1/((1-x)^2) ?

The integral from 0.43 to x of 1/((1-x)^2) is 1/(1-x)-1.75438…

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integral from 0.43 to x of 1/((1-x)^2)

\int_{0.43}^{x} \frac{1}{( 1 - x ) ^{2}} d x

\frac{1}{1 - x} - 1.75438 \dots

Solution steps

\int_{0.43}^{x} \frac{1}{( 1 - x ) ^{2}} d x

Apply u-substitution

: \int_{0.57}^{1 - x} - \frac{1}{u ^{2}} d u

= \int_{0.57}^{1 - x} - \frac{1}{u ^{2}} d u

Take the constant out:

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

= - \int_{0.57}^{1 - x} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{0.57}^{1 - x}

= - [- \frac{1}{u}]_{0.57}^{1 - x}

Compute the boundaries:

- \frac{1}{1 - x} + 1.75438 \dots

= - (- \frac{1}{1 - x} + 1.75438 \dots)

Simplify

= \frac{1}{1 - x} - 1.75438 \dots

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What is the integral from 0.43 to x of 1/((1-x)^2) ?
The integral from 0.43 to x of 1/((1-x)^2) is 1/(1-x)-1.75438…