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

Lösung

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

x^{2} + x + 3 - \frac{1}{2} ln \frac{2}{11} x + \frac{1}{11} + \frac{4 x ^{2} + 4 x + 12}{11} + C

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} + x + 3 : (x + \frac{1}{2})^{2} + \frac{11}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{11}{4}} d x

Wende U-Substitution an

= \int \frac{2 u - 1}{4 u ^{2} + 11} d u

Multipliziere aus

\frac{2 u - 1}{4 u ^{2} + 11} : \frac{2 u}{4 u ^{2} + 11} - \frac{1}{4 u ^{2} + 11}

= \int \frac{2 u}{4 u ^{2} + 11} - \frac{1}{4 u ^{2} + 11} d u

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{2 u}{4 u ^{2} + 11} d u - \int \frac{1}{4 u ^{2} + 11} d u

\int \frac{2 u}{4 u ^{2} + 11} d u = \frac{1}{2} 4 u^{2} + 11

\int \frac{1}{4 u ^{2} + 11} d u = \frac{1}{2} ln \frac{2}{11} u + \frac{1}{11} (11 + 4 u^{2})

= \frac{1}{2} 4 u^{2} + 11 - \frac{1}{2} ln \frac{2}{11} u + \frac{1}{11} (11 + 4 u^{2})

Setze in

u = x + \frac{1}{2}

ein

= \frac{1}{2} 4 (x + \frac{1}{2})^{2} + 11 - \frac{1}{2} ln \frac{2}{11} (x + \frac{1}{2}) + \frac{1}{11} (11 + 4 (x + \frac{1}{2})^{2})

Vereinfache

\frac{1}{2} 4 (x + \frac{1}{2})^{2} + 11 - \frac{1}{2} ln \frac{2}{11} (x + \frac{1}{2}) + \frac{1}{11} (11 + 4 (x + \frac{1}{2})^{2}) : x^{2} + x + 3 - \frac{1}{2} ln \frac{2}{11} x + \frac{1}{11} + \frac{4 x ^{2} + 4 x + 12}{11}

= x^{2} + x + 3 - \frac{1}{2} ln \frac{2}{11} x + \frac{1}{11} + \frac{4 x ^{2} + 4 x + 12}{11}

Füge eine Konstante zur Lösung hinzu

= x^{2} + x + 3 - \frac{1}{2} ln \frac{2}{11} x + \frac{1}{11} + \frac{4 x ^{2} + 4 x + 12}{11} + C

\frac{\partial}{\partial x} (\frac{x - 2 y}{y ^{2} + 5 x ^{2}})

d er i v a t i v e \frac{1}{2} (6 t + 4)^{- \frac{1}{2}}

\int_{0}^{7} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

\frac{d}{d x} (2 x^{3} - 6 x + 5)

x \to 3 lim (\frac{( x )}{x - 3})