What is the integral from 0 to 0.9 of 1/((1-x)^2) ?

The integral from 0 to 0.9 of 1/((1-x)^2) is 9

Popular >

integral from 0 to 0.9 of 1/((1-x)^2)

\int_{0}^{0.9} \frac{1}{( 1 - x ) ^{2}} d x

9

Solution steps

\int_{0}^{0.9} \frac{1}{( 1 - x ) ^{2}} d x

Apply u-substitution

: \int_{1}^{0.1} - \frac{1}{u ^{2}} d u

= \int_{1}^{0.1} - \frac{1}{u ^{2}} d u

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

= - \int_{0.1}^{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_{0.1}^{1} \frac{1}{u ^{2}} d u)

Apply the Power Rule

: [- \frac{1}{u}]_{0.1}^{1}

= - (- [- \frac{1}{u}]_{0.1}^{1})

Simplify

= [- \frac{1}{u}]_{0.1}^{1}

Compute the boundaries:

9

= 9

\int_{0}^{\frac{π}{2}} (cos (x))^{5} d x

\int_{1}^{e} x ln (x) d x

\int_{2}^{6} \frac{x}{16} d x

\int_{- 7}^{0} (5 + 49 - x^{2}) d x

\int_{0}^{3} 10 t (75 - t^{2}) d t

What is the integral from 0 to 0.9 of 1/((1-x)^2) ?
The integral from 0 to 0.9 of 1/((1-x)^2) is 9