integral from x to x+4 of 1/(t^2+2)

Lösung

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

\frac{1}{2} (arctan (\frac{x + 4}{2}) - arctan (\frac{x}{2}))

Schritte zur Lösung

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

Wende integrale Substitution an

= \int_{\frac{x}{2}}^{\frac{x + 4}{2}} \frac{1}{2 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{\frac{x}{2}}^{\frac{x + 4}{2}} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{2} [arctan (u)]_{\frac{x}{2}}^{\frac{x + 4}{2}}

Berechne die Grenzen:

arctan (\frac{x + 4}{2}) - arctan (\frac{x}{2})

= \frac{1}{2} (arctan (\frac{x + 4}{2}) - arctan (\frac{x}{2}))

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