integral of (1-4x)/(sqrt(9x^2+6x-3))

Lösung

\int \frac{1 - 4 x}{9 x ^{2} + 6 x - 3} d x

- \frac{4}{9} 9 x^{2} + 6 x - 3 + \frac{7}{9} ln (\frac{9 x ^{2} + 6 x - 3 + 3 x + 1}{2}) + C

Schritte zur Lösung

\int \frac{1 - 4 x}{9 x ^{2} + 6 x - 3} d x

Vervollständige das Quadrat

9 x^{2} + 6 x - 3 : 9 (x + \frac{1}{3})^{2} - 4

= \int \frac{1 - 4 x}{9 ( x + \frac{1}{3} ) ^{2} - 4} d x

Wende U-Substitution an

= \int \frac{1 - 4 ( u - \frac{1}{3} )}{9 u ^{2} - 4} d u

Multipliziere aus

\frac{1 - 4 ( u - \frac{1}{3} )}{9 u ^{2} - 4} : - \frac{4 u}{9 u ^{2} - 4} + \frac{7}{3 9 u ^{2} - 4}

= \int - \frac{4 u}{9 u ^{2} - 4} + \frac{7}{3 9 u ^{2} - 4} 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

= - \int \frac{4 u}{9 u ^{2} - 4} d u + \int \frac{7}{3 9 u ^{2} - 4} d u

\int \frac{4 u}{9 u ^{2} - 4} d u = \frac{4}{9} 9 u^{2} - 4

\int \frac{7}{3 9 u ^{2} - 4} d u = \frac{7}{9} ln (\frac{9 u ^{2} - 4 + 3 u}{2})

= - \frac{4}{9} 9 u^{2} - 4 + \frac{7}{9} ln (\frac{9 u ^{2} - 4 + 3 u}{2})

Setze in

u = x + \frac{1}{3}

ein

= - \frac{4}{9} 9 (x + \frac{1}{3})^{2} - 4 + \frac{7}{9} ln \frac{9 ( x + \frac{1}{3} ) ^{2} - 4 + 3 ( x + \frac{1}{3} )}{2}

Vereinfache

- \frac{4}{9} 9 (x + \frac{1}{3})^{2} - 4 + \frac{7}{9} ln \frac{9 ( x + \frac{1}{3} ) ^{2} - 4 + 3 ( x + \frac{1}{3} )}{2} : - \frac{4}{9} 9 x^{2} + 6 x - 3 + \frac{7}{9} ln (\frac{9 x ^{2} + 6 x - 3 + 3 x + 1}{2})

= - \frac{4}{9} 9 x^{2} + 6 x - 3 + \frac{7}{9} ln (\frac{9 x ^{2} + 6 x - 3 + 3 x + 1}{2})

Füge eine Konstante zur Lösung hinzu

= - \frac{4}{9} 9 x^{2} + 6 x - 3 + \frac{7}{9} ln (\frac{9 x ^{2} + 6 x - 3 + 3 x + 1}{2}) + C

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