integral from 0 to 2 of 1/((x^2+4)^2)

Lösung

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

\frac{π + 2}{64}

Dezimale

0.08033 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{0}^{1} \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Trigonometrische Substitution anwenden

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 v )}{2} d v

Entferne die Konstante:

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

= \frac{1}{8} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 v) d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d v + \int_{0}^{\frac{π}{4}} cos (2 v) d v)

\int_{0}^{\frac{π}{4}} 1 d v = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 v) d v = \frac{1}{2}

= \frac{1}{8} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Vereinfache

\frac{1}{8} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{π + 2}{64}

= \frac{π + 2}{64}