integral of (x+8)/(x^2+x+2)

Lösung

\int \frac{x + 8}{x ^{2} + x + 2} d x

\frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{15}{7} arctan (\frac{2 x + 1}{7}) + C

Schritte zur Lösung

\int \frac{x + 8}{x ^{2} + x + 2} d x

Wende U-Substitution an

= \int \frac{u}{u ^{2} - 15 u + 58} d u

Vervollständige das Quadrat

u^{2} - 15 u + 58 : (u - \frac{15}{2})^{2} + \frac{7}{4}

= \int \frac{u}{( u - \frac{15}{2} ) ^{2} + \frac{7}{4}} d u

Wende U-Substitution an

= \int \frac{2 ( 2 v + 15 )}{4 v ^{2} + 7} d v

Entferne die Konstante:

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

= 2 \cdot \int \frac{2 v + 15}{4 v ^{2} + 7} d v

Multipliziere aus

\frac{2 v + 15}{4 v ^{2} + 7} : \frac{2 v}{4 v ^{2} + 7} + \frac{15}{4 v ^{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{2 v}{4 v ^{2} + 7} d v + \int \frac{15}{4 v ^{2} + 7} d v)

\int \frac{2 v}{4 v ^{2} + 7} d v = \frac{1}{4} ln 4 v^{2} + 7

\int \frac{15}{4 v ^{2} + 7} d v = \frac{15}{2 7} arctan (\frac{2}{7} v)

= 2 (\frac{1}{4} ln 4 v^{2} + 7 + \frac{15}{2 7} arctan (\frac{2}{7} v))

Ersetze zurück

= 2 (\frac{1}{4} ln 4 (x + 8 - \frac{15}{2})^{2} + 7 + \frac{15}{2 7} arctan (\frac{2}{7} (x + 8 - \frac{15}{2})))

Vereinfache

2 (\frac{1}{4} ln 4 (x + 8 - \frac{15}{2})^{2} + 7 + \frac{15}{2 7} arctan (\frac{2}{7} (x + 8 - \frac{15}{2}))) : \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{15}{7} arctan (\frac{2 x + 1}{7})

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{15}{7} arctan (\frac{2 x + 1}{7})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{15}{7} arctan (\frac{2 x + 1}{7}) + C

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