integral of ((4x-5))/((x-2)(x+3))

Lösung

\int \frac{( 4 x - 5 )}{( x - 2 ) ( x + 3 )} d x

\frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{4 x - 5}{( x - 2 ) ( x + 3 )} d x

Wende U-Substitution an

= \int \frac{4 u - 17}{u ( u - 5 )} d u

Ermittle den Partialbruch von

\frac{4 u - 17}{u ( u - 5 )} : \frac{17}{5 u} + \frac{3}{5 ( u - 5 )}

= \int \frac{17}{5 u} + \frac{3}{5 ( u - 5 )} d u

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{17}{5 u} d u + \int \frac{3}{5 ( u - 5 )} d u

\int \frac{17}{5 u} d u = \frac{17}{5} ln ∣ u ∣

\int \frac{3}{5 ( u - 5 )} d u = \frac{3}{5} ln ∣ u - 5 ∣

= \frac{17}{5} ln ∣ u ∣ + \frac{3}{5} ln ∣ u - 5 ∣

Setze in

u = x + 3

ein

= \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x + 3 - 5 ∣

Vereinfache

\frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x + 3 - 5 ∣ : \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣

= \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣ + C

\int \frac{y}{x ( x + y )} d x

x y^{^{'}} + y = e^{x}, y (1) = 6

\int \frac{9}{6 + 9 x} d x

\int (2^{x} + e^{π}) d x

d er i v a t i v e e^{7}