integral of (8x-3)/(x^2+x+1)

Lösung

\int \frac{8 x - 3}{x ^{2} + x + 1} d x

16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

Schritte zur Lösung

\int \frac{8 x - 3}{x ^{2} + x + 1} d x

Multipliziere aus

\frac{8 x - 3}{x ^{2} + x + 1} : \frac{8 x}{x ^{2} + x + 1} - \frac{3}{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{8 x}{x ^{2} + x + 1} d x - \int \frac{3}{x ^{2} + x + 1} d x

\int \frac{8 x}{x ^{2} + x + 1} d x = 16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3}))

\int \frac{3}{x ^{2} + x + 1} d x = 23 arctan (\frac{1}{3} (2 x + 1))

= 16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1))

Füge eine Konstante zur Lösung hinzu

= 16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

x \to 0 + lim (\frac{∣ x ∣}{x})

t an g e n t f (x) = 7 + 4 x^{2} - 2 x^{3}, (2, 7)

y^{^{'}} = (\frac{2 sin ( x )}{sin ( y )}), y (0) = \frac{π}{2}

\int \frac{x}{( 1 - x ) ^{4}} d x

x \to 3 lim (x^{2} - 2 x)