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

Lösung

\int_{0}^{π} \frac{1}{a + b x} d x

\frac{ln ∣ a + πb ∣ - ln ∣ a ∣}{b}

Schritte zur Lösung

\int_{0}^{π} \frac{1}{a + b x} d x

Wende U-Substitution an

= \int_{a}^{a + bπ} \frac{1}{b u} d u

Entferne die Konstante:

\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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{b} [ln ∣ u ∣]_{a}^{a + bπ}

Berechne die Grenzen:

ln ∣ a + πb ∣ - ln ∣ a ∣

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

Vereinfache

= \frac{ln ∣ a + πb ∣ - ln ∣ a ∣}{b}

\int_{x}^{5 x} e^{3 t} d t

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