integral from 1 to 8 of (4x)/(x^2+6)

Lösung

\int_{1}^{8} \frac{4 x}{x ^{2} + 6} d x

2 ln (10)

Dezimale

4.60517 \dots

Schritte zur Lösung

\int_{1}^{8} \frac{4 x}{x ^{2} + 6} d x

Entferne die Konstante:

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

= 4 \cdot \int_{1}^{8} \frac{x}{x ^{2} + 6} d x

Wende U-Substitution an

= 4 \cdot \int_{7}^{70} \frac{1}{2 u} d u

Entferne die Konstante:

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

= 4 \cdot \frac{1}{2} \cdot \int_{7}^{70} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{70}

Vereinfache

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{70} : 2 [ln ∣ u ∣]_{7}^{70}

= 2 [ln ∣ u ∣]_{7}^{70}

Berechne die Grenzen:

ln (70) - ln (7)

= 2 (ln (70) - ln (7))

Vereinfache

= 2 ln (10)

\int_{2}^{4} (5 x^{3} + 2 x^{2} - 3 x - 8) d x

\int_{3}^{6} 15 ln (x + 5) d x

\int_{1}^{4} (2 x^{2} - 4 x + 5) d x

\int_{0}^{\frac{π}{4}} tan (t) d t

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