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Beliebt Rechnen >

integral from 0 to 2 of 1/(8+2t^2)

Lösung

\int_{0}^{2} \frac{1}{8 + 2 t ^{2}} d t

Lösung

\frac{π}{16}

+1

Dezimale

0.19634 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{8 + 2 t ^{2}} d t

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{4 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{4} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{16}

Graph

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\int \frac{2 x + 1}{( x ^{2} + 4 ) ^{2} ( x + 1 ) ^{3}} d x

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