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

Lösung

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

2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7})) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

2 x^{2} + x + 1 : 2 (x + \frac{1}{4})^{2} + \frac{7}{8}

= \int \frac{x}{2 ( x + \frac{1}{4} ) ^{2} + \frac{7}{8}} d x

Wende U-Substitution an

= \int \frac{2 ( 4 u - 1 )}{16 u ^{2} + 7} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{4 u - 1}{16 u ^{2} + 7} d u

Multipliziere aus

\frac{4 u - 1}{16 u ^{2} + 7} : \frac{4 u}{16 u ^{2} + 7} - \frac{1}{16 u ^{2} + 7}

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{4 u}{16 u ^{2} + 7} d u - \int \frac{1}{16 u ^{2} + 7} d u)

\int \frac{4 u}{16 u ^{2} + 7} d u = \frac{1}{8} ln 16 u^{2} + 7

\int \frac{1}{16 u ^{2} + 7} d u = \frac{1}{4 7} arctan (\frac{4}{7} u)

= 2 (\frac{1}{8} ln 16 u^{2} + 7 - \frac{1}{4 7} arctan (\frac{4}{7} u))

Setze in

u = x + \frac{1}{4}

ein

= 2 (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 7 - \frac{1}{4 7} arctan (\frac{4}{7} (x + \frac{1}{4})))

Vereinfache

2 (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 7 - \frac{1}{4 7} arctan (\frac{4}{7} (x + \frac{1}{4}))) : 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7}))

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7}))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 8 - \frac{1}{4 7} arctan (\frac{4 x + 1}{7})) + C

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