integral of (x^2+x)/(x^2+x+1)

Lösung

\int \frac{x ^{2} + x}{x ^{2} + x + 1} d x

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

Schritte zur Lösung

\int \frac{x ^{2} + x}{x ^{2} + x + 1} d x

Multipliziere aus

\frac{x ^{2} + x}{x ^{2} + x + 1} : \frac{x ^{2}}{x ^{2} + x + 1} + \frac{x}{x ^{2} + x + 1}

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{x ^{2}}{x ^{2} + x + 1} d x + \int \frac{x}{x ^{2} + x + 1} d x

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e y = e^{- 11 x}

n = 1 \sum \infty n^{3}

\frac{d x}{d t} = 3 t^{2}

\int \frac{1}{1 - 3 t} d t

y^{^{'}} + 7 y = t + e^{- 5 t}