integral from 0 to 5 of x/(a+bx^2)

Lösung

\int_{0}^{5} \frac{x}{a + b x ^{2}} d x

\frac{ln ∣ a + 25 b ∣ - ln ∣ a ∣}{2 b}

Schritte zur Lösung

\int_{0}^{5} \frac{x}{a + b x ^{2}} d x

Wende U-Substitution an

= \int_{a}^{a + 25 b} \frac{1}{2 b u} d u

Entferne die Konstante:

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

= \frac{1}{2 b} \cdot \int_{a}^{a + 25 b} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2 b} [ln ∣ u ∣]_{a}^{a + 25 b}

Berechne die Grenzen:

ln ∣ a + 25 b ∣ - ln ∣ a ∣

= \frac{1}{2 b} (ln ∣ a + 25 b ∣ - ln ∣ a ∣)

Vereinfache

= \frac{ln ∣ a + 25 b ∣ - ln ∣ a ∣}{2 b}

\int_{0}^{1} x^{9} (1 + x^{10})^{9} d x

\int_{- 1}^{1} \frac{x ^{3} + 1}{x ^{2} + 1} d x

\int_{\frac{1}{2}}^{2} 5 - \frac{5}{x ^{2}} d x

\int_{0}^{1} (x^{2} - 1)^{2} d x

\int_{2}^{4} 3 x^{2} - 2 d x