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

Lösung

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

ln \frac{1}{8} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2 2} (x + 2)) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Vervollständige das Quadrat

x^{2} + 4 x + 12 : (x + 2)^{2} + 8

= 2 \cdot \int \frac{x}{( x + 2 ) ^{2} + 8} d x

Wende U-Substitution an

= 2 \cdot \int \frac{u - 2}{u ^{2} + 8} d u

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

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

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

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

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

Ersetze zurück

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

Vereinfache

2 \cdot \frac{1}{2} (\frac{1}{2} ln (\frac{x + 2}{2 2})^{2} + 1 - arctan (\frac{x + 2}{2 2})) : ln \frac{1}{8} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2 2} (x + 2))

= ln \frac{1}{8} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2 2} (x + 2))

Füge eine Konstante zur Lösung hinzu

= ln \frac{1}{8} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2 2} (x + 2)) + C

\int \frac{cos ( x )}{1 + sin ^{2} ( x )} d x

\frac{d}{d x} (ln (\frac{3 + e ^{x}}{3 - e ^{x}}))

\frac{d}{d x} (a x^{4} + b x^{3} + c x^{2})

d er i v a t i v e (3 x - 1)^{7} (2 x + 1)^{5}

\frac{d}{d y} (e^{2 y})