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

The integral from 0 to 0.9 of 1/(1-x) is 2.30258…

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

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

2.30258 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0.1}^{1})

Simplify

= [ln ∣ u ∣]_{0.1}^{1}

Compute the boundaries:

2.30258 \dots

= 2.30258 \dots

8 y^{^{''}} + 2 y^{^{'}} - y = 0

sin (x), cos (x), [0, \frac{π}{2}]

\frac{d y}{d x} = 2 x e^{4 y}

y^{^{''}} - 2 y^{^{'}} + y = - 6 - 3 x + \frac{2 e ^{x}}{x}

\frac{d}{d x} (ln (x^{2} e^{x^{2 + 1}}))

What is the integral from 0 to 0.9 of 1/(1-x) ?
The integral from 0 to 0.9 of 1/(1-x) is 2.30258…