What is the integral from 0 to 0.8 of 1/((1-x)^2) ?

The integral from 0 to 0.8 of 1/((1-x)^2) is 4

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integral from 0 to 0.8 of 1/((1-x)^2)

\int_{0}^{0.8} \frac{1}{( 1 - x ) ^{2}} d x

4

Solution steps

\int_{0}^{0.8} \frac{1}{( 1 - x ) ^{2}} d x

Apply u-substitution

: \int_{1}^{0.2} - \frac{1}{u ^{2}} d u

= \int_{1}^{0.2} - \frac{1}{u ^{2}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0.2}^{1} - \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_{0.2}^{1} \frac{1}{u ^{2}} d u)

Apply the Power Rule

: [- \frac{1}{u}]_{0.2}^{1}

= - (- [- \frac{1}{u}]_{0.2}^{1})

Simplify

= [- \frac{1}{u}]_{0.2}^{1}

Compute the boundaries:

4

= 4

\int_{2}^{4} \frac{2 ( 1 + x )}{27} d x

\int_{1}^{2} - x^{2} + 4 d x

\int_{0}^{2} e^{- x^{2}} x d x

\int_{1}^{3} \frac{x ^{3}}{x} d x

What is the integral from 0 to 0.8 of 1/((1-x)^2) ?
The integral from 0 to 0.8 of 1/((1-x)^2) is 4