integral of (x^2-3x-7)/((2x+3)(x+1))

Lösung

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 )} d x

\frac{1}{4} (2 x + ln ∣ 2 x + 3 ∣ - 12 ln ∣ 2 x + 2 ∣ + 2) + C

Schritte zur Lösung

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 )} d x

Wende U-Substitution an

= \int \frac{u ^{2} - 12 u - 1}{4 u ( u - 1 )} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{u ^{2} - 12 u - 1}{u ( u - 1 )} d u

Multipliziere aus

\frac{u ^{2} - 12 u - 1}{u ( u - 1 )} : \frac{u ^{2}}{u ( u - 1 )} - \frac{12 u}{u ( u - 1 )} - \frac{1}{u ( u - 1 )}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int \frac{u ^{2}}{u ( u - 1 )} d u - \int \frac{12 u}{u ( u - 1 )} d u - \int \frac{1}{u ( u - 1 )} d u)

\int \frac{u ^{2}}{u ( u - 1 )} d u = u - 1 + ln ∣ u - 1 ∣

\int \frac{12 u}{u ( u - 1 )} d u = 12 ln ∣ u - 1 ∣

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

= \frac{1}{4} (u - 1 + ln ∣ u - 1 ∣ - 12 ln ∣ u - 1 ∣ - (- ln ∣ u ∣ + ln ∣ u - 1 ∣))

Setze in

u = 2 x + 3

ein

= \frac{1}{4} (2 x + 3 - 1 + ln ∣ 2 x + 3 - 1 ∣ - 12 ln ∣ 2 x + 3 - 1 ∣ - (- ln ∣ 2 x + 3 ∣ + ln ∣ 2 x + 3 - 1 ∣))

Vereinfache

\frac{1}{4} (2 x + 3 - 1 + ln ∣ 2 x + 3 - 1 ∣ - 12 ln ∣ 2 x + 3 - 1 ∣ - (- ln ∣ 2 x + 3 ∣ + ln ∣ 2 x + 3 - 1 ∣)) : \frac{1}{4} (2 x + ln ∣ 2 x + 3 ∣ - 12 ln ∣ 2 x + 2 ∣ + 2)

= \frac{1}{4} (2 x + ln ∣ 2 x + 3 ∣ - 12 ln ∣ 2 x + 2 ∣ + 2)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 x + ln ∣ 2 x + 3 ∣ - 12 ln ∣ 2 x + 2 ∣ + 2) + C

\int_{1}^{e^{3}} \frac{1}{x ( 1 + ln ( x ) )} d x

y^{^{''}} + 25 y = 4 csc^{2} (5 t)

\frac{\partial}{\partial x} (3 x cos (y))

x \to 0 lim (\frac{1}{cos ( 8 x )})

\frac{d}{d x} (- 11 (4 x^{3} - 11)^{13})