integral of (7x+1)/(3x^2+6x)

Lösung

\int \frac{7 x + 1}{3 x ^{2} + 6 x} d x

\frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{7 x + 1}{3 x ^{2} + 6 x} d x

Multipliziere aus

\frac{7 x + 1}{3 x ^{2} + 6 x} : \frac{7 x}{3 x ^{2} + 6 x} + \frac{1}{3 x ^{2} + 6 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{7 x}{3 x ^{2} + 6 x} d x + \int \frac{1}{3 x ^{2} + 6 x} d x

\int \frac{7 x}{3 x ^{2} + 6 x} d x = \frac{7}{3} ln ∣ x + 2 ∣

\int \frac{1}{3 x ^{2} + 6 x} d x = - \frac{1}{6} (ln ∣ x + 2 ∣ - ln ∣ x ∣)

= \frac{7}{3} ln ∣ x + 2 ∣ - \frac{1}{6} (ln ∣ x + 2 ∣ - ln ∣ x ∣)

Vereinfache

\frac{7}{3} ln ∣ x + 2 ∣ - \frac{1}{6} (ln ∣ x + 2 ∣ - ln ∣ x ∣) : \frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣

= \frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{13}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x ∣ + C

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