integral of 1/((2x+1)(x+2))

Lösung

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

\frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣ + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{1}{u ( u + 3 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} 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{1}{3 u} d u - \int \frac{1}{3 ( u + 3 )} d u

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= \frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣

Setze in

u = 2 x + 1

ein

= \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 1 + 3 ∣

Vereinfache

\frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 1 + 3 ∣ : \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣

= \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣ + C

\int \frac{5 x ^{2} - x - 4}{x ^{2} ( x + 1 )} d x

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

\int sin (3 y) cos (5 y) d y

\int 5 t^{4} - t^{2} (10 t^{3} - 5 t) d t

\int \frac{1}{x ^{2} 4 x ^{2} + 1} d x