integral from 8 to 11 of y/(y^2-4y-5)

Lösung

\int_{8}^{11} \frac{y}{y ^{2} - 4 y - 5} d y

\frac{ln ( \frac{4}{3} ) + 5 ln ( 2 )}{6}

Dezimale

0.62556 \dots

Schritte zur Lösung

\int_{8}^{11} \frac{y}{y ^{2} - 4 y - 5} d y

Ermittle den Partialbruch von

\frac{y}{y ^{2} - 4 y - 5} : \frac{1}{6 ( y + 1 )} + \frac{5}{6 ( y - 5 )}

= \int_{8}^{11} \frac{1}{6 ( y + 1 )} + \frac{5}{6 ( y - 5 )} 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_{8}^{11} \frac{1}{6 ( y + 1 )} d y + \int_{8}^{11} \frac{5}{6 ( y - 5 )} d y

\int_{8}^{11} \frac{1}{6 ( y + 1 )} d y = \frac{1}{6} ln (\frac{4}{3})

\int_{8}^{11} \frac{5}{6 ( y - 5 )} d y = \frac{5}{6} ln (2)

= \frac{1}{6} ln (\frac{4}{3}) + \frac{5}{6} ln (2)

Vereinfache

\frac{1}{6} ln (\frac{4}{3}) + \frac{5}{6} ln (2) : \frac{ln ( \frac{4}{3} ) + 5 ln ( 2 )}{6}

= \frac{ln ( \frac{4}{3} ) + 5 ln ( 2 )}{6}

d er i v a t i v e y = 3 x^{2} + e^{2 x} + π^{2}

a re a y = \frac{3}{x}, y = 9 x, y = \frac{1}{9} x

\int_{3}^{4} x x - 3 d x

t an g e n t f (x) = \frac{1}{x + 9}, at x = - 7

\frac{\partial}{\partial x} (3 y e^{4 x y} + 8 x)