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

Lösung

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

\frac{3}{2} ln 36 x^{2} + 36 x + 36 - \frac{5}{3} arctan (\frac{2 x + 1}{3}) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{3 u}{u ^{2} + 5 u + 13} d u

Entferne die Konstante:

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

= 3 \cdot \int \frac{u}{u ^{2} + 5 u + 13} d u

Vervollständige das Quadrat

u^{2} + 5 u + 13 : (u + \frac{5}{2})^{2} + \frac{27}{4}

= 3 \cdot \int \frac{u}{( u + \frac{5}{2} ) ^{2} + \frac{27}{4}} d u

Wende U-Substitution an

= 3 \cdot \int \frac{2 ( 2 v - 5 )}{4 v ^{2} + 27} d v

Entferne die Konstante:

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

= 3 \cdot 2 \cdot \int \frac{2 v - 5}{4 v ^{2} + 27} d v

Multipliziere aus

\frac{2 v - 5}{4 v ^{2} + 27} : \frac{2 v}{4 v ^{2} + 27} - \frac{5}{4 v ^{2} + 27}

Wende die Summenregel an:

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

= 3 \cdot 2 (\int \frac{2 v}{4 v ^{2} + 27} d v - \int \frac{5}{4 v ^{2} + 27} d v)

\int \frac{2 v}{4 v ^{2} + 27} d v = \frac{1}{4} ln 4 v^{2} + 27

\int \frac{5}{4 v ^{2} + 27} d v = \frac{5}{6 3} arctan (\frac{2}{3 3} v)

= 3 \cdot 2 (\frac{1}{4} ln 4 v^{2} + 27 - \frac{5}{6 3} arctan (\frac{2}{3 3} v))

Ersetze zurück

= 3 \cdot 2 (\frac{1}{4} ln 4 (3 x - 1 + \frac{5}{2})^{2} + 27 - \frac{5}{6 3} arctan (\frac{2}{3 3} (3 x - 1 + \frac{5}{2})))

Vereinfache

3 \cdot 2 (\frac{1}{4} ln 4 (3 x - 1 + \frac{5}{2})^{2} + 27 - \frac{5}{6 3} arctan (\frac{2}{3 3} (3 x - 1 + \frac{5}{2}))) : \frac{3}{2} ln 36 x^{2} + 36 x + 36 - \frac{5}{3} arctan (\frac{2 x + 1}{3})

= \frac{3}{2} ln 36 x^{2} + 36 x + 36 - \frac{5}{3} arctan (\frac{2 x + 1}{3})

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} ln 36 x^{2} + 36 x + 36 - \frac{5}{3} arctan (\frac{2 x + 1}{3}) + C

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