What is the integral from a to p of b/(x(1-x)) ?

The integral from a to p of b/(x(1-x)) is b(ln|p|-ln|a|-ln|p-1|+ln|a-1|)

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integral from a to p of b/(x(1-x))

\int_{a}^{p} \frac{b}{x ( 1 - x )} d x

b (ln ∣ p ∣ - ln ∣ a ∣ - ln ∣ p - 1 ∣ + ln ∣ a - 1 ∣)

Solution steps

\int_{a}^{p} \frac{b}{x ( 1 - x )} d x

Take the constant out:

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

= b \cdot \int_{a}^{p} \frac{1}{x ( 1 - x )} d x

Take the partial fraction of

\frac{1}{x ( 1 - x )} : \frac{1}{x} - \frac{1}{x - 1}

= b \cdot \int_{a}^{p} \frac{1}{x} - \frac{1}{x - 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= b (\int_{a}^{p} \frac{1}{x} d x - \int_{a}^{p} \frac{1}{x - 1} d x)

\int_{a}^{p} \frac{1}{x} d x = ln ∣ p ∣ - ln ∣ a ∣

\int_{a}^{p} \frac{1}{x - 1} d x = ln ∣ p - 1 ∣ - ln ∣ a - 1 ∣

= b (ln ∣ p ∣ - ln ∣ a ∣ - (ln ∣ p - 1 ∣ - ln ∣ a - 1 ∣))

Simplify

= b (ln ∣ p ∣ - ln ∣ a ∣ - ln ∣ p - 1 ∣ + ln ∣ a - 1 ∣)

n = 0 \sum \infty \frac{e ^{n}}{n}

\frac{d y}{d x} = 0.5 y + x

What is the integral from a to p of b/(x(1-x)) ?
The integral from a to p of b/(x(1-x)) is b(ln|p|-ln|a|-ln|p-1|+ln|a-1|)