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

Lösung

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

4 (- \frac{3}{16} - 4 x^{2} - x + 1 - \frac{19}{64} arcsin (\frac{8 x + 1}{17})) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

1 - x - 4 x^{2} : - 4 (x + \frac{1}{8})^{2} + \frac{17}{16}

= \int \frac{3 x - 2}{- 4 ( x + \frac{1}{8} ) ^{2} + \frac{17}{16}} d x

Wende U-Substitution an

= \int \frac{4 ( 3 ( u - \frac{1}{8} ) - 2 )}{- 64 u ^{2} + 17} d u

Entferne die Konstante:

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

= 4 \cdot \int \frac{3 ( u - \frac{1}{8} ) - 2}{- 64 u ^{2} + 17} d u

Multipliziere aus

\frac{3 ( u - \frac{1}{8} ) - 2}{- 64 u ^{2} + 17} : \frac{3 u}{- 64 u ^{2} + 17} - \frac{19}{8 - 64 u ^{2} + 17}

= 4 \cdot \int \frac{3 u}{- 64 u ^{2} + 17} - \frac{19}{8 - 64 u ^{2} + 17} 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

= 4 (\int \frac{3 u}{- 64 u ^{2} + 17} d u - \int \frac{19}{8 - 64 u ^{2} + 17} d u)

\int \frac{3 u}{- 64 u ^{2} + 17} d u = - \frac{3}{64} - 64 u^{2} + 17

\int \frac{19}{8 - 64 u ^{2} + 17} d u = \frac{19}{64} arcsin (\frac{8}{17} u)

= 4 (- \frac{3}{64} - 64 u^{2} + 17 - \frac{19}{64} arcsin (\frac{8}{17} u))

Setze in

u = x + \frac{1}{8}

ein

= 4 - \frac{3}{64} - 64 (x + \frac{1}{8})^{2} + 17 - \frac{19}{64} arcsin (\frac{8}{17} (x + \frac{1}{8}))

Vereinfache

4 - \frac{3}{64} - 64 (x + \frac{1}{8})^{2} + 17 - \frac{19}{64} arcsin (\frac{8}{17} (x + \frac{1}{8})) : 4 (- \frac{3}{16} - 4 x^{2} - x + 1 - \frac{19}{64} arcsin (\frac{8 x + 1}{17}))

= 4 (- \frac{3}{16} - 4 x^{2} - x + 1 - \frac{19}{64} arcsin (\frac{8 x + 1}{17}))

Füge eine Konstante zur Lösung hinzu

= 4 (- \frac{3}{16} - 4 x^{2} - x + 1 - \frac{19}{64} arcsin (\frac{8 x + 1}{17})) + C

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