integral of (x^2)/((x-4)(x-5)^2)

Lösung

\int \frac{x ^{2}}{( x - 4 ) ( x - 5 ) ^{2}} d x

16 ln ∣ x - 4 ∣ - 15 ln ∣ x - 5 ∣ - \frac{25}{x - 5} + C

Schritte zur Lösung

\int \frac{x ^{2}}{( x - 4 ) ( x - 5 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{x ^{2}}{( x - 4 ) ( x - 5 ) ^{2}} : \frac{16}{x - 4} - \frac{15}{x - 5} + \frac{25}{( x - 5 ) ^{2}}

= \int \frac{16}{x - 4} - \frac{15}{x - 5} + \frac{25}{( x - 5 ) ^{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{16}{x - 4} d x - \int \frac{15}{x - 5} d x + \int \frac{25}{( x - 5 ) ^{2}} d x

\int \frac{16}{x - 4} d x = 16 ln ∣ x - 4 ∣

\int \frac{15}{x - 5} d x = 15 ln ∣ x - 5 ∣

\int \frac{25}{( x - 5 ) ^{2}} d x = - \frac{25}{x - 5}

= 16 ln ∣ x - 4 ∣ - 15 ln ∣ x - 5 ∣ - \frac{25}{x - 5}

Füge eine Konstante zur Lösung hinzu

= 16 ln ∣ x - 4 ∣ - 15 ln ∣ x - 5 ∣ - \frac{25}{x - 5} + C

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

y^{^{''}} + 9 y = cos (5 t)

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

\int \frac{( sin ( 3 x ) - cos ( 2 x ) )}{4} d x

x \to - 3 lim (\frac{x ^{2} - 4}{x - 2})