integral of (3x)/(x^2+4x+5)

Lösung

\int \frac{3 x}{x ^{2} + 4 x + 5} d x

3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + C

Schritte zur Lösung

\int \frac{3 x}{x ^{2} + 4 x + 5} 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 + 5} d x

Vervollständige das Quadrat

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Wende U-Substitution an

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

Multipliziere aus

\frac{u - 2}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} - \frac{2}{u ^{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{u}{u ^{2} + 1} d u - \int \frac{2}{u ^{2} + 1} d u)

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

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

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

Setze in

u = x + 2

ein

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

Vereinfache

3 (\frac{1}{2} ln (x + 2)^{2} + 1 - 2 arctan (x + 2)) : 3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2))

= 3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2))

Füge eine Konstante zur Lösung hinzu

= 3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + C

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