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

Lösung

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

\frac{1}{6 3} (36 x^{2} - 12 x + 12 + ln \frac{6}{11} x - \frac{1}{11} + \frac{36 x ^{2} - 12 x + 12}{11}) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

1 - x + 3 x^{2} : 3 (x - \frac{1}{6})^{2} + \frac{11}{12}

= \int \frac{x}{3 ( x - \frac{1}{6} ) ^{2} + \frac{11}{12}} d x

Wende U-Substitution an

= \int \frac{6 u + 1}{3 36 u ^{2} + 11} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{6 u + 1}{36 u ^{2} + 11} d u

Multipliziere aus

\frac{6 u + 1}{36 u ^{2} + 11} : \frac{6 u}{36 u ^{2} + 11} + \frac{1}{36 u ^{2} + 11}

= \frac{1}{3} \cdot \int \frac{6 u}{36 u ^{2} + 11} + \frac{1}{36 u ^{2} + 11} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{6 u}{36 u ^{2} + 11} d u + \int \frac{1}{36 u ^{2} + 11} d u)

\int \frac{6 u}{36 u ^{2} + 11} d u = \frac{1}{6} 36 u^{2} + 11

\int \frac{1}{36 u ^{2} + 11} d u = \frac{1}{6} ln \frac{6}{11} u + \frac{1}{11} (11 + 36 u^{2})

= \frac{1}{3} (\frac{1}{6} 36 u^{2} + 11 + \frac{1}{6} ln \frac{6}{11} u + \frac{1}{11} (11 + 36 u^{2}))

Setze in

u = x - \frac{1}{6}

ein

= \frac{1}{3} \frac{1}{6} 36 (x - \frac{1}{6})^{2} + 11 + \frac{1}{6} ln \frac{6}{11} (x - \frac{1}{6}) + \frac{1}{11} (11 + 36 (x - \frac{1}{6})^{2})

Vereinfache

\frac{1}{3} \frac{1}{6} 36 (x - \frac{1}{6})^{2} + 11 + \frac{1}{6} ln \frac{6}{11} (x - \frac{1}{6}) + \frac{1}{11} (11 + 36 (x - \frac{1}{6})^{2}) : \frac{1}{6 3} (36 x^{2} - 12 x + 12 + ln \frac{6}{11} x - \frac{1}{11} + \frac{36 x ^{2} - 12 x + 12}{11})

= \frac{1}{6 3} (36 x^{2} - 12 x + 12 + ln \frac{6}{11} x - \frac{1}{11} + \frac{36 x ^{2} - 12 x + 12}{11})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6 3} (36 x^{2} - 12 x + 12 + ln \frac{6}{11} x - \frac{1}{11} + \frac{36 x ^{2} - 12 x + 12}{11}) + C

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