integral from 1 to 3 of y/((y+1)(y^2+2))

Lösung

\int_{1}^{3} \frac{y}{( y + 1 ) ( y ^{2} + 2 )} d y

\frac{- 2 ln ( 2 ) + ln ( \frac{11}{2} ) - ln ( \frac{3}{2} ) + 2 2 arctan ( \frac{3 2}{2} ) - 2 2 arctan ( \frac{2}{2} )}{6}

Dezimale

0.22817 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{y}{( y + 1 ) ( y ^{2} + 2 )} d y

Ermittle den Partialbruch von

\frac{y}{( y + 1 ) ( y ^{2} + 2 )} : - \frac{1}{3 ( y + 1 )} + \frac{y + 2}{3 ( y ^{2} + 2 )}

= \int_{1}^{3} - \frac{1}{3 ( y + 1 )} + \frac{y + 2}{3 ( y ^{2} + 2 )} d y

Wende die Summenregel an:

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

= - \int_{1}^{3} \frac{1}{3 ( y + 1 )} d y + \int_{1}^{3} \frac{y + 2}{3 ( y ^{2} + 2 )} d y

\int_{1}^{3} \frac{1}{3 ( y + 1 )} d y = \frac{1}{3} ln (2)

\int_{1}^{3} \frac{y + 2}{3 ( y ^{2} + 2 )} d y = \frac{ln ( \frac{11}{2} ) - ln ( \frac{3}{2} ) + 2 2 arctan ( \frac{3 2}{2} ) - 2 2 arctan ( \frac{2}{2} )}{6}

= - \frac{1}{3} ln (2) + \frac{ln ( \frac{11}{2} ) - ln ( \frac{3}{2} ) + 2 2 arctan ( \frac{3 2}{2} ) - 2 2 arctan ( \frac{2}{2} )}{6}

Vereinfache

- \frac{1}{3} ln (2) + \frac{ln ( \frac{11}{2} ) - ln ( \frac{3}{2} ) + 2 2 arctan ( \frac{3 2}{2} ) - 2 2 arctan ( \frac{2}{2} )}{6} : \frac{- 2 ln ( 2 ) + ln ( \frac{11}{2} ) - ln ( \frac{3}{2} ) + 2 2 arctan ( \frac{3 2}{2} ) - 2 2 arctan ( \frac{2}{2} )}{6}

= \frac{- 2 ln ( 2 ) + ln ( \frac{11}{2} ) - ln ( \frac{3}{2} ) + 2 2 arctan ( \frac{3 2}{2} ) - 2 2 arctan ( \frac{2}{2} )}{6}

\int_{1}^{7} t^{2} - 2 t - 15 d t

\int_{2}^{3} \frac{1}{x ^{2} - x} d x

\int_{- 4}^{0} x + 4 d x

\int_{1}^{9} \frac{3}{x} d x

\int_{1}^{3} 2^{3 x} d x