integral of 4/(x^2+x)

Lösung

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

4 (ln ∣ x ∣ - ln ∣ x + 1 ∣) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

= 4 \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}

= 4 \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

= 4 (\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 ∣

= 4 (ln ∣ x ∣ - ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= 4 (ln ∣ x ∣ - ln ∣ x + 1 ∣) + C

\int \frac{x ^{2} + 6 x + 5}{x + 3} d x

\int - 3 e^{- 2 t} d t

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

\int (3 t^{- 4}) (2 + 4 t^{- 3})^{- 7} d t

\int [ln (ln (x)) + \frac{1}{ln ( x )}] d x