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

Lösung

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

2 (2 - 1 - e (ln (2 + e) - ln (1 + e)))

Dezimale

0.25421 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1 + e}^{2 + e} \frac{2 ( u - e )}{u} d u

Entferne die Konstante:

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

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

Multipliziere aus

\frac{u - e}{u} : 1 - \frac{e}{u}

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{1 + e}^{2 + e} 1 d u - \int_{1 + e}^{2 + e} \frac{e}{u} d u)

\int_{1 + e}^{2 + e} 1 d u = 2 - 1

\int_{1 + e}^{2 + e} \frac{e}{u} d u = e (ln (2 + e) - ln (1 + e))

= 2 (2 - 1 - e (ln (2 + e) - ln (1 + e)))

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