integral of 2/(x^2+x)

Lösung

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

2 (ln ∣ x ∣ - ln ∣ x + 1 ∣) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

= 2 \cdot \int \frac{1}{x ^{2} + x} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} + x} : \frac{1}{x} - \frac{1}{x + 1}

= 2 \cdot \int \frac{1}{x} - \frac{1}{x + 1} 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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣) + C

\frac{d}{d x} (\frac{6 x ^{2} + 8 x + 4}{x})

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

\frac{d x}{d y} = sin (x + y)

y^{^{'}} = - a y

\frac{d}{d x} (x^{a} y^{1 - a})