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

Lösung

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

\frac{π + 2}{8}

Dezimale

0.64269 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Entferne die Konstante:

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

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

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

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

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

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

Vereinfache

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

= \frac{π + 2}{8}

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

d er i v a t i v e (arctan (8 x))^{2}

\int_{1}^{e^{2}} ln (x) d x

y^{^{''}} + 14 y^{^{'}} + 58 y = 0

d er i v a t i v e \frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}