integral of (x^2-2)/(x^2+x)

Lösung

\int \frac{x ^{2} - 2}{x ^{2} + x} d x

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

Schritte zur Lösung

\int \frac{x ^{2} - 2}{x ^{2} + x} d x

Multipliziere aus

\frac{x ^{2} - 2}{x ^{2} + x} : \frac{x ^{2}}{x ^{2} + x} - \frac{2}{x ^{2} + 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{x ^{2}}{x ^{2} + x} d x - \int \frac{2}{x ^{2} + x} d x

\int \frac{x ^{2}}{x ^{2} + x} d x = \frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1}

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

= \frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣)

Vereinfache

\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣) : \frac{x ^{2}}{x + 1} + ln ∣ x + 1 ∣ - \frac{1}{x + 1} - 2 ln ∣ x ∣

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

Füge eine Konstante zur Lösung hinzu

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

(x + 2) \frac{d y}{d x} + y = ln (x), y (1) = 20

a re a x, x^{2}, 0, 1

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

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

\int (3 t + 3)^{2.2} d t