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

Lösung

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

\frac{- 2 ln ( 2 ) + π}{8}

Dezimale

0.21941 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Wende die Summenregel an:

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

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

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

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

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

Vereinfache

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

= \frac{- 2 ln ( 2 ) + π}{8}

\int x (x - 5)^{2} d x

d er i v a t i v e f (x) = \frac{x ^{3}}{24} + \frac{2}{x}

\int_{1}^{2} \frac{1}{4 - t ^{2}} d t

t an g e n t - 3 x^{2} + 2 x y + 2 y^{3} = 125, (3, 4)

\frac{d}{d x} y = ln (10 x^{2} - 8)