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Beliebt Rechnen >

integral of (x+1)/(3x^2+6x-5)

Lösung

\int \frac{x + 1}{3 x ^{2} + 6 x - 5} d x

Lösung

\frac{1}{6} ln 3 x^{2} + 6 x - 5 + C

Schritte zur Lösung

\int \frac{x + 1}{3 x ^{2} + 6 x - 5} d x

Multipliziere aus

\frac{x + 1}{3 x ^{2} + 6 x - 5} : \frac{x}{3 x ^{2} + 6 x - 5} + \frac{1}{3 x ^{2} + 6 x - 5}

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}{3 x ^{2} + 6 x - 5} d x + \int \frac{1}{3 x ^{2} + 6 x - 5} d x

\int \frac{x}{3 x ^{2} + 6 x - 5} d x = \frac{1}{6} ln 3 x^{2} + 6 x - 5 + \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1)

\int \frac{1}{3 x ^{2} + 6 x - 5} d x = - \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1)

= \frac{1}{6} ln 3 x^{2} + 6 x - 5 + \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1) - \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1)

Addiere gleiche Elemente:

\frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1) - \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1) = 0

= \frac{1}{6} ln 3 x^{2} + 6 x - 5

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln 3 x^{2} + 6 x - 5 + C

Graph

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