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

The integral from 0 to 1 of (ax)/(x^2-px) is a(ln|1-p|-ln|p|)

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

\int_{0}^{1} \frac{a x}{x ^{2} - p x} d x

a (ln ∣ 1 - p ∣ - ln ∣ p ∣)

Solution steps

\int_{0}^{1} \frac{a x}{x ^{2} - p x} d x

Take the constant out:

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

= a \cdot \int_{0}^{1} \frac{x}{x ^{2} - p x} d x

Simplify

\frac{x}{x ^{2} - p x} : \frac{1}{x - p}

= a \cdot \int_{0}^{1} \frac{1}{x - p} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a [ln ∣ u ∣]_{- p}^{1 - p}

Compute the boundaries:

ln ∣ 1 - p ∣ - ln ∣ p ∣

= a (ln ∣ 1 - p ∣ - ln ∣ p ∣)

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What is the integral from 0 to 1 of (ax)/(x^2-px) ?
The integral from 0 to 1 of (ax)/(x^2-px) is a(ln|1-p|-ln|p|)