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

Lösung

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

2 (- \frac{1}{4} ln 4 x^{2} - 12 x - 4 - \frac{1}{4 13} (ln \frac{2 x - 3}{13} + 1 - ln \frac{2 x - 3}{13} - 1)) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} - 3 x - 1 : (x - \frac{3}{2})^{2} - \frac{13}{4}

= \int \frac{2 - x}{( x - \frac{3}{2} ) ^{2} - \frac{13}{4}} d x

Wende U-Substitution an

= \int \frac{2 ( - 2 u + 1 )}{4 u ^{2} - 13} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{- 2 u + 1}{4 u ^{2} - 13} d u

Multipliziere aus

\frac{- 2 u + 1}{4 u ^{2} - 13} : - \frac{2 u}{4 u ^{2} - 13} + \frac{1}{4 u ^{2} - 13}

Wende die Summenregel an:

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

= 2 (- \int \frac{2 u}{4 u ^{2} - 13} d u + \int \frac{1}{4 u ^{2} - 13} d u)

\int \frac{2 u}{4 u ^{2} - 13} d u = \frac{1}{4} ln 4 u^{2} - 13

\int \frac{1}{4 u ^{2} - 13} d u = - \frac{1}{4 13} (ln \frac{2}{13} u + 1 - ln \frac{2}{13} u - 1)

= 2 (- \frac{1}{4} ln 4 u^{2} - 13 - \frac{1}{4 13} (ln \frac{2}{13} u + 1 - ln \frac{2}{13} u - 1))

Setze in

u = x - \frac{3}{2}

ein

= 2 (- \frac{1}{4} ln 4 (x - \frac{3}{2})^{2} - 13 - \frac{1}{4 13} (ln \frac{2}{13} (x - \frac{3}{2}) + 1 - ln \frac{2}{13} (x - \frac{3}{2}) - 1))

Vereinfache

2 (- \frac{1}{4} ln 4 (x - \frac{3}{2})^{2} - 13 - \frac{1}{4 13} (ln \frac{2}{13} (x - \frac{3}{2}) + 1 - ln \frac{2}{13} (x - \frac{3}{2}) - 1)) : 2 (- \frac{1}{4} ln 4 x^{2} - 12 x - 4 - \frac{1}{4 13} (ln \frac{2 x - 3}{13} + 1 - ln \frac{2 x - 3}{13} - 1))

= 2 (- \frac{1}{4} ln 4 x^{2} - 12 x - 4 - \frac{1}{4 13} (ln \frac{2 x - 3}{13} + 1 - ln \frac{2 x - 3}{13} - 1))

Füge eine Konstante zur Lösung hinzu

= 2 (- \frac{1}{4} ln 4 x^{2} - 12 x - 4 - \frac{1}{4 13} (ln \frac{2 x - 3}{13} + 1 - ln \frac{2 x - 3}{13} - 1)) + C

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