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

Lösung

\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{2 x - 1}{3}) + x + C

Schritte zur Lösung

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

\frac{x ^{2}}{x ^{2} - x + 1} = \frac{x - 1}{x ^{2} - x + 1} + 1

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

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

\int \frac{x - 1}{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})

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (A \cdot e^{x})

d er i v a t i v e f (x) = \frac{x ^{5}}{5} - \frac{4 x ^{3}}{3} + 4 x

t an g e n t f (x) = 7 e^{- 0.03 x}, at x = 9

d er i v a t i v e y = \frac{( 3 x - 9 )}{( 2 x + 8 )}

\int (x^{3} + x) e^{- 4 x} d x