integral of x/(x^2-5x+4)

Lösung

\int \frac{x}{x ^{2} - 5 x + 4} d x

2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x)) + C

Schritte zur Lösung

\int \frac{x}{x ^{2} - 5 x + 4} d x

Vervollständige das Quadrat

x^{2} - 5 x + 4 : (x - \frac{5}{2})^{2} - \frac{9}{4}

= \int \frac{x}{( x - \frac{5}{2} ) ^{2} - \frac{9}{4}} d x

Wende U-Substitution an

= \int \frac{2 ( 2 u + 5 )}{4 u ^{2} - 9} 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 + 5}{4 u ^{2} - 9} d u

Multipliziere aus

\frac{2 u + 5}{4 u ^{2} - 9} : \frac{2 u}{4 u ^{2} - 9} + \frac{5}{4 u ^{2} - 9}

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} - 9} d u + \int \frac{5}{4 u ^{2} - 9} d u)

\int \frac{2 u}{4 u ^{2} - 9} d u = \frac{1}{4} ln 4 u^{2} - 9

\int \frac{5}{4 u ^{2} - 9} d u = - \frac{5}{12} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 9 - \frac{5}{12} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1))

Setze in

u = x - \frac{5}{2}

ein

= 2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 9 - \frac{5}{12} (ln \frac{2}{3} (x - \frac{5}{2}) + 1 - ln \frac{2}{3} (x - \frac{5}{2}) - 1))

Vereinfache

2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 9 - \frac{5}{12} (ln \frac{2}{3} (x - \frac{5}{2}) + 1 - ln \frac{2}{3} (x - \frac{5}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x))

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x)) + C

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