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

Lösung

integral

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

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

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Vervollständige das Quadrat

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

= 3 \cdot \int \frac{x}{( x + 2 ) ^{2} + 2} d x

Wende U-Substitution an

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

Wende integrale Substitution an

= 3 \cdot \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

= 3 (\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)

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{( 2 x )}{( 4 + 3 x ) ^{2}}

in t e g r a l \frac{( 1 - cos ( x ) ) ^{2}}{sin ^{2} ( x )}

in t e g r a l 3^{- x}

in t e g r a l \frac{arctan ( x )}{x ^{2}}

in t e g r a l \frac{1}{( x - 2 ) ^{3} ( x + 4 ) ^{4}}