integral from 0 to 1 of (2x)/(x^2+1)

Lösung

\int_{0}^{1} \frac{2 x}{x ^{2} + 1} d x

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{2 x}{x ^{2} + 1} d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} \frac{x}{x ^{2} + 1} d x

Wende U-Substitution an

= 2 \cdot \int_{1}^{2} \frac{1}{2 u} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Vereinfache

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : [ln ∣ u ∣]_{1}^{2}

= [ln ∣ u ∣]_{1}^{2}

Berechne die Grenzen:

ln (2)

= ln (2)

in v erse l a pl a ce \frac{s}{1 + s}

\int \frac{x}{3 x ^{2} + 1} d x

5 \frac{d y}{d x} + 45 y = 9

\frac{d}{d x} (x^{2} cos (y) y^{^{'}})

\frac{d}{d x} (1 - e^{2 x})