What is the integral from 0 to pi of 1/(a+bx) ?

The integral from 0 to pi of 1/(a+bx) is (ln|a+pib|-ln|a|)/b

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integral from 0 to pi of 1/(a+bx)

\int_{0}^{π} \frac{1}{a + b x} d x

\frac{ln ∣ a + πb ∣ - ln ∣ a ∣}{b}

Solution steps

\int_{0}^{π} \frac{1}{a + b x} d x

Apply u-substitution

: \int_{a}^{a + bπ} \frac{1}{b u} d u

= \int_{a}^{a + bπ} \frac{1}{b u} d u

Take the constant out:

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

= \frac{1}{b} \cdot \int_{a}^{a + bπ} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{b} [ln ∣ u ∣]_{a}^{a + bπ}

Compute the boundaries:

ln ∣ a + πb ∣ - ln ∣ a ∣

= \frac{1}{b} (ln ∣ a + πb ∣ - ln ∣ a ∣)

Simplify

= \frac{ln ∣ a + πb ∣ - ln ∣ a ∣}{b}

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What is the integral from 0 to pi of 1/(a+bx) ?
The integral from 0 to pi of 1/(a+bx) is (ln|a+pib|-ln|a|)/b