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

Lösung

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

4 - 22

Dezimale

1.17157 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [\frac{2}{1 + x} x - \int - \frac{x}{2 ( x + 1 ) ^{\frac{3}{2}}} d x]_{0}^{1}

\int - \frac{x}{2 ( x + 1 ) ^{\frac{3}{2}}} d x = - \frac{2 ( x + 2 )}{x + 1}

= [\frac{2}{1 + x} x - (- \frac{2 ( x + 2 )}{x + 1})]_{0}^{1}

Vereinfache

[\frac{2}{1 + x} x - (- \frac{2 ( x + 2 )}{x + 1})]_{0}^{1} : [22 1 + x]_{0}^{1}

= [22 1 + x]_{0}^{1}

Berechne die Grenzen:

4 - 22

= 4 - 22

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