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

Lösung

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

\frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{x}{x ^{2} + 6 x + 1} d x = \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22

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

= \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1) + C

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