integral of (x+2)/(x^2-4x+8)

Lösung

\int \frac{x + 2}{x ^{2} - 4 x + 8} d x

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

Schritte zur Lösung

\int \frac{x + 2}{x ^{2} - 4 x + 8} d x

Vervollständige das Quadrat

x^{2} - 4 x + 8 : (x - 2)^{2} + 4

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

Wende U-Substitution an

= \int \frac{u + 4}{u ^{2} + 4} d u

Wende integrale Substitution an

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

Multipliziere aus

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

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

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

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

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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