integral of (2x-1)/(5x^2-x+3)

Lösung

\int \frac{2 x - 1}{5 x ^{2} - x + 3} d x

\frac{1}{5} ln \frac{( 10 x - 1 ) ^{2}}{59} + 1 - \frac{8}{5 59} arctan (\frac{1}{59} (10 x - 1)) + C

Schritte zur Lösung

\int \frac{2 x - 1}{5 x ^{2} - x + 3} d x

Wende U-Substitution an

= \int \frac{2 u}{5 u ^{2} + 8 u + 15} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{u}{5 u ^{2} + 8 u + 15} d u

Vervollständige das Quadrat

5 u^{2} + 8 u + 15 : 5 (u + \frac{4}{5})^{2} + \frac{59}{5}

= 2 \cdot \int \frac{u}{5 ( u + \frac{4}{5} ) ^{2} + \frac{59}{5}} d u

Wende U-Substitution an

= 2 \cdot \int \frac{5 v - 4}{25 v ^{2} + 59} d v

Wende integrale Substitution an

= 2 \cdot \int \frac{59 w - 4}{5 59 ( w ^{2} + 1 )} d w

Entferne die Konstante:

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

= 2 \cdot \frac{1}{5 59} \cdot \int \frac{59 w - 4}{w ^{2} + 1} d w

Multipliziere aus

\frac{59 w - 4}{w ^{2} + 1} : \frac{59 w}{w ^{2} + 1} - \frac{4}{w ^{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

= 2 \cdot \frac{1}{5 59} (\int \frac{59 w}{w ^{2} + 1} d w - \int \frac{4}{w ^{2} + 1} d w)

\int \frac{59 w}{w ^{2} + 1} d w = \frac{59}{2} ln w^{2} + 1

\int \frac{4}{w ^{2} + 1} d w = 4 arctan (w)

= 2 \cdot \frac{1}{5 59} (\frac{59}{2} ln w^{2} + 1 - 4 arctan (w))

Ersetze zurück

= 2 \cdot \frac{1}{5 59} (\frac{59}{2} ln (\frac{5}{59} (2 x - 1 + \frac{4}{5}))^{2} + 1 - 4 arctan (\frac{5}{59} (2 x - 1 + \frac{4}{5})))

Vereinfache

2 \cdot \frac{1}{5 59} (\frac{59}{2} ln (\frac{5}{59} (2 x - 1 + \frac{4}{5}))^{2} + 1 - 4 arctan (\frac{5}{59} (2 x - 1 + \frac{4}{5}))) : \frac{1}{5} ln \frac{( 10 x - 1 ) ^{2}}{59} + 1 - \frac{8}{5 59} arctan (\frac{1}{59} (10 x - 1))

= \frac{1}{5} ln \frac{( 10 x - 1 ) ^{2}}{59} + 1 - \frac{8}{5 59} arctan (\frac{1}{59} (10 x - 1))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln \frac{( 10 x - 1 ) ^{2}}{59} + 1 - \frac{8}{5 59} arctan (\frac{1}{59} (10 x - 1)) + C