integral of (5x^2+6x+9)/((x-3)^2(x+1)^2)

Lösung

\int \frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} d x

- \frac{9}{2 ( x - 3 )} - \frac{1}{2 ( x + 1 )} + C

Schritte zur Lösung

\int \frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} : \frac{9}{2 ( x - 3 ) ^{2}} + \frac{1}{2 ( x + 1 ) ^{2}}

= \int \frac{9}{2 ( x - 3 ) ^{2}} + \frac{1}{2 ( x + 1 ) ^{2}} 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 \frac{9}{2 ( x - 3 ) ^{2}} d x + \int \frac{1}{2 ( x + 1 ) ^{2}} d x

\int \frac{9}{2 ( x - 3 ) ^{2}} d x = - \frac{9}{2 ( x - 3 )}

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

= - \frac{9}{2 ( x - 3 )} - \frac{1}{2 ( x + 1 )}

Füge eine Konstante zur Lösung hinzu

= - \frac{9}{2 ( x - 3 )} - \frac{1}{2 ( x + 1 )} + C

\frac{d}{d x} (\frac{x ^{6}}{1 + x ^{12}})

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

\frac{\partial}{\partial y} (x^{2} + y^{2} - 25)

\int - 0.04 t d t

\frac{d}{d x} (- \frac{2}{9} x^{- \frac{5}{3}})