integral of (8x-5)/(3x^2-5x+2)

Lösung

\int \frac{8 x - 5}{3 x ^{2} - 5 x + 2} d x

\frac{4}{3} ln 36 x^{2} - 60 x + 24 - \frac{20}{3} (ln ∣ 6 x - 4 ∣ - ln ∣ 6 x - 6 ∣) + 5 ln ∣ 6 x - 4 ∣ - 5 ln ∣ 6 x - 6 ∣ + C

Schritte zur Lösung

\int \frac{8 x - 5}{3 x ^{2} - 5 x + 2} d x

Vervollständige das Quadrat

3 x^{2} - 5 x + 2 : 3 (x - \frac{5}{6})^{2} - \frac{1}{12}

= \int \frac{8 x - 5}{3 ( x - \frac{5}{6} ) ^{2} - \frac{1}{12}} d x

Wende U-Substitution an

= \int \frac{12 ( 8 ( u + \frac{5}{6} ) - 5 )}{36 u ^{2} - 1} d u

Entferne die Konstante:

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

= 12 \cdot \int \frac{8 ( u + \frac{5}{6} ) - 5}{36 u ^{2} - 1} d u

Multipliziere aus

\frac{8 ( u + \frac{5}{6} ) - 5}{36 u ^{2} - 1} : \frac{8 ( u + \frac{5}{6} )}{36 u ^{2} - 1} - \frac{5}{36 u ^{2} - 1}

Wende die Summenregel an:

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

= 12 (\int \frac{8 ( u + \frac{5}{6} )}{36 u ^{2} - 1} d u - \int \frac{5}{36 u ^{2} - 1} d u)

\int \frac{8 ( u + \frac{5}{6} )}{36 u ^{2} - 1} d u = 8 (\frac{1}{72} ln 36 u^{2} - 1 - \frac{5}{72} (ln ∣ 6 u + 1 ∣ - ln ∣ 6 u - 1 ∣))

\int \frac{5}{36 u ^{2} - 1} d u = - \frac{5}{12} (ln ∣ 6 u + 1 ∣ - ln ∣ 6 u - 1 ∣)

= 12 (8 (\frac{1}{72} ln 36 u^{2} - 1 - \frac{5}{72} (ln ∣ 6 u + 1 ∣ - ln ∣ 6 u - 1 ∣)) - (- \frac{5}{12} (ln ∣ 6 u + 1 ∣ - ln ∣ 6 u - 1 ∣)))

Setze in

u = x - \frac{5}{6}

ein

= 12 (8 (\frac{1}{72} ln 36 (x - \frac{5}{6})^{2} - 1 - \frac{5}{72} (ln 6 (x - \frac{5}{6}) + 1 - ln 6 (x - \frac{5}{6}) - 1)) - (- \frac{5}{12} (ln 6 (x - \frac{5}{6}) + 1 - ln 6 (x - \frac{5}{6}) - 1)))

Vereinfache

12 (8 (\frac{1}{72} ln 36 (x - \frac{5}{6})^{2} - 1 - \frac{5}{72} (ln 6 (x - \frac{5}{6}) + 1 - ln 6 (x - \frac{5}{6}) - 1)) - (- \frac{5}{12} (ln 6 (x - \frac{5}{6}) + 1 - ln 6 (x - \frac{5}{6}) - 1))) : \frac{4}{3} ln 36 x^{2} - 60 x + 24 - \frac{20}{3} (ln ∣ 6 x - 4 ∣ - ln ∣ 6 x - 6 ∣) + 5 ln ∣ 6 x - 4 ∣ - 5 ln ∣ 6 x - 6 ∣

= \frac{4}{3} ln 36 x^{2} - 60 x + 24 - \frac{20}{3} (ln ∣ 6 x - 4 ∣ - ln ∣ 6 x - 6 ∣) + 5 ln ∣ 6 x - 4 ∣ - 5 ln ∣ 6 x - 6 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{4}{3} ln 36 x^{2} - 60 x + 24 - \frac{20}{3} (ln ∣ 6 x - 4 ∣ - ln ∣ 6 x - 6 ∣) + 5 ln ∣ 6 x - 4 ∣ - 5 ln ∣ 6 x - 6 ∣ + C

\int \frac{3 a}{4 t} d t

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in t e g r a l \frac{1}{u ^{3}}

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