integral from 1 to 3 of (3x^2)/(x^3+1)

Lösung

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

ln (14)

Dezimale

2.63905 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 3 \cdot \int_{2}^{28} \frac{1}{3 u} d u

Entferne die Konstante:

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

= 3 \cdot \frac{1}{3} \cdot \int_{2}^{28} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{2}^{28}

Vereinfache

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{2}^{28} : [ln ∣ u ∣]_{2}^{28}

= [ln ∣ u ∣]_{2}^{28}

Berechne die Grenzen:

ln (28) - ln (2)

= ln (28) - ln (2)

Vereinfache

= ln (14)

\int_{1 5^{\circ}}^{3 0^{\circ}} x^{2} sin (5) x^{3} d x

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

\int_{- 1}^{- 2} f (t)

\int_{0}^{1} [x^{\frac{1}{n}} - x] d x

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