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

Lösung

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

\frac{arctan ( \frac{2}{y} ) - arctan ( \frac{1}{y} )}{y}

Schritte zur Lösung

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

= \frac{1}{y} [arctan (u)]_{\frac{1}{y}}^{\frac{2}{y}}

Berechne die Grenzen:

arctan (\frac{2}{y}) - arctan (\frac{1}{y})

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

Vereinfache

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

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