integral of (-x+4)/(12(x^2-2x+4))

Lösung

\int \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x

\frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1))) + C

Schritte zur Lösung

\int \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x

Entferne die Konstante:

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

= \frac{1}{12} \cdot \int \frac{- x + 4}{x ^{2} - 2 x + 4} d x

Vervollständige das Quadrat

x^{2} - 2 x + 4 : (x - 1)^{2} + 3

= \frac{1}{12} \cdot \int \frac{- x + 4}{( x - 1 ) ^{2} + 3} d x

Wende U-Substitution an

= \frac{1}{12} \cdot \int \frac{- u + 3}{u ^{2} + 3} d u

Wende integrale Substitution an

= \frac{1}{12} \cdot \int \frac{- v + 3}{v ^{2} + 1} d v

Multipliziere aus

\frac{- v + 3}{v ^{2} + 1} : - \frac{v}{v ^{2} + 1} + \frac{3}{v ^{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

= \frac{1}{12} (- \int \frac{v}{v ^{2} + 1} d v + \int \frac{3}{v ^{2} + 1} d v)

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

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= \frac{1}{12} (- \frac{1}{2} ln v^{2} + 1 + 3 arctan (v))

Ersetze zurück

= \frac{1}{12} (- \frac{1}{2} ln (\frac{x - 1}{3})^{2} + 1 + 3 arctan (\frac{x - 1}{3}))

Vereinfache

\frac{1}{12} (- \frac{1}{2} ln (\frac{x - 1}{3})^{2} + 1 + 3 arctan (\frac{x - 1}{3})) : \frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1)))

= \frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1)))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1))) + C

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