integral of (6x-5)/(sqrt(3x^2+4x+1))

Lösung

\int \frac{6 x - 5}{3 x ^{2} + 4 x + 1} d x

\frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2})) + C

Schritte zur Lösung

\int \frac{6 x - 5}{3 x ^{2} + 4 x + 1} d x

Wende U-Substitution an

= \int \frac{u}{3 u ^{2} + 18 u + 77} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{u}{u ^{2} + 18 u + 77} d u

Vervollständige das Quadrat

u^{2} + 18 u + 77 : (u + 9)^{2} - 4

= \frac{1}{3} \cdot \int \frac{u}{( u + 9 ) ^{2} - 4} d u

Wende U-Substitution an

= \frac{1}{3} \cdot \int \frac{v - 9}{v ^{2} - 4} d v

Multipliziere aus

\frac{v - 9}{v ^{2} - 4} : \frac{v}{v ^{2} - 4} - \frac{9}{v ^{2} - 4}

= \frac{1}{3} \cdot \int \frac{v}{v ^{2} - 4} - \frac{9}{v ^{2} - 4} d v

Wende die Summenregel an:

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

= \frac{1}{3} (\int \frac{v}{v ^{2} - 4} d v - \int \frac{9}{v ^{2} - 4} d v)

\int \frac{v}{v ^{2} - 4} d v = v^{2} - 4

\int \frac{9}{v ^{2} - 4} d v = 9 ln (\frac{1}{2} v^{2} - 4 + v)

= \frac{1}{3} (v^{2} - 4 - 9 ln (\frac{1}{2} v^{2} - 4 + v))

Ersetze zurück

= \frac{1}{3} ((6 x - 5 + 9)^{2} - 4 - 9 ln (\frac{1}{2} (6 x - 5 + 9)^{2} - 4 + 6 x - 5 + 9))

Vereinfache

\frac{1}{3} ((6 x - 5 + 9)^{2} - 4 - 9 ln (\frac{1}{2} (6 x - 5 + 9)^{2} - 4 + 6 x - 5 + 9)) : \frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2}))

= \frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2})) + C

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