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

Lösung

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

\frac{π - 2 arctan ( \frac{2}{2} )}{2 2}

Dezimale

0.67551 \dots

Schritte zur Lösung

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

Berechne das unbestimmte Integral:

\int \frac{1}{x ^{2} + 2} d x = \frac{1}{2} arctan (\frac{x}{2}) + C

Berechne die Grenzen:

\int_{1}^{\infty} \frac{1}{x ^{2} + 2} d x = \frac{π}{2 2} - \frac{1}{2} arctan (\frac{1}{2})

= \frac{π}{2 2} - \frac{1}{2} arctan (\frac{1}{2})

Vereinfache

= \frac{π - 2 arctan ( \frac{2}{2} )}{2 2}

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\int_{0}^{\frac{1}{2}} \frac{5}{1 + 4 x ^{2}} d x

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