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integral of (x^2)/(x^3+4x^2+6x)

Lösung

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

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

Schritte zur Lösung

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

Faktorisiere

\frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x}

= \int \frac{x ^{2}}{x ( x ^{2} + 4 x + 6 )} d x

Streiche die gemeinsamen Faktoren:

x

= \int \frac{x}{x ^{2} + 4 x + 6} d x

Vervollständige das Quadrat

x^{2} + 4 x + 6 : (x + 2)^{2} + 2

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

Wende U-Substitution an

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

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

Füge eine Konstante zur Lösung hinzu

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

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