integral from 0 to 1 of (2x)/(x^2+4)

Lösung

\int_{0}^{1} \frac{2 x}{x ^{2} + 4} d x

ln (5) - 2 ln (2)

Dezimale

0.22314 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{2 x}{x ^{2} + 4} 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} + 4} d x

Wende U-Substitution an

= 2 \cdot \int_{4}^{5} \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_{4}^{5} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{4}^{5}

Vereinfache

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{4}^{5} : [ln ∣ u ∣]_{4}^{5}

= [ln ∣ u ∣]_{4}^{5}

Berechne die Grenzen:

ln (5) - 2 ln (2)

= ln (5) - 2 ln (2)

\int_{- 2}^{3} (x^{2} - 2) d x

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

\int_{1}^{3} 1 + x d x

\int_{- 2}^{2} 10 x^{9} + 25 x^{17} d x

\int_{- 4}^{- 2} y^{2} + \frac{1}{y ^{3}} d y