integral of x/(x^2+3x+9)

Lösung

\int \frac{x}{x ^{2} + 3 x + 9} d x

\frac{1}{2} ln 4 x^{2} + 12 x + 36 - \frac{1}{3} arctan (\frac{2}{3 3} x + \frac{1}{3}) + C

Schritte zur Lösung

\int \frac{x}{x ^{2} + 3 x + 9} d x

Vervollständige das Quadrat

x^{2} + 3 x + 9 : (x + \frac{3}{2})^{2} + \frac{27}{4}

= \int \frac{x}{( x + \frac{3}{2} ) ^{2} + \frac{27}{4}} d x

Wende U-Substitution an

= \int \frac{2 ( 2 u - 3 )}{4 u ^{2} + 27} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{2 u - 3}{4 u ^{2} + 27} d u

Multipliziere aus

\frac{2 u - 3}{4 u ^{2} + 27} : \frac{2 u}{4 u ^{2} + 27} - \frac{3}{4 u ^{2} + 27}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{2 u}{4 u ^{2} + 27} d u - \int \frac{3}{4 u ^{2} + 27} d u)

\int \frac{2 u}{4 u ^{2} + 27} d u = \frac{1}{4} ln 4 u^{2} + 27

\int \frac{3}{4 u ^{2} + 27} d u = \frac{1}{2 3} arctan (\frac{2}{3 3} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 27 - \frac{1}{2 3} arctan (\frac{2}{3 3} u))

Setze in

u = x + \frac{3}{2}

ein

= 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 27 - \frac{1}{2 3} arctan (\frac{2}{3 3} (x + \frac{3}{2})))

Vereinfache

2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 27 - \frac{1}{2 3} arctan (\frac{2}{3 3} (x + \frac{3}{2}))) : \frac{1}{2} ln 4 x^{2} + 12 x + 36 - \frac{1}{3} arctan (\frac{2}{3 3} x + \frac{1}{3})

= \frac{1}{2} ln 4 x^{2} + 12 x + 36 - \frac{1}{3} arctan (\frac{2}{3 3} x + \frac{1}{3})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln 4 x^{2} + 12 x + 36 - \frac{1}{3} arctan (\frac{2}{3 3} x + \frac{1}{3}) + C

\int 2 x^{- 3} d x

\int_{0}^{4} \frac{2 t}{( t - 5 ) ^{2}} d t

x \to 0 lim (ln (1 + x))

\frac{\partial}{\partial z} (8 x^{\frac{y}{z}})

l a pl a ce t r an s f or m 3 (t - 2)^{2} + 13 (t - 2) + 17