integral from 2 to 9 of 1/(sqrt(8x+9))

Lösung

\int_{2}^{9} \frac{1}{8 x + 9} d x

1

Schritte zur Lösung

\int_{2}^{9} \frac{1}{8 x + 9} d x

Wende U-Substitution an

= \int_{25}^{81} \frac{1}{8 u} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{25}^{81} \frac{1}{u} d u

Wende die Potenzregel an

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{25}^{81}

Wende Radikal Regel an:

a^{\frac{1}{n}} = n a

= \frac{1}{8} [2 u]_{25}^{81}

Berechne die Grenzen:

8

= \frac{1}{8} \cdot 8

Vereinfache

= 1

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