integral of (-6x^4+6)/(x^2+1)

Lösung

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

- 2 x^{3} + 6 x + C

Schritte zur Lösung

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

Multipliziere aus

\frac{- 6 x ^{4} + 6}{x ^{2} + 1} : - \frac{6 x ^{4}}{x ^{2} + 1} + \frac{6}{x ^{2} + 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{6 x ^{4}}{x ^{2} + 1} d x + \int \frac{6}{x ^{2} + 1} d x

\int \frac{6 x ^{4}}{x ^{2} + 1} d x = 6 (\frac{x ^{3}}{3} - x + arctan (x))

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= - 6 (\frac{x ^{3}}{3} - x + arctan (x)) + 6 arctan (x)

Vereinfache

- 6 (\frac{x ^{3}}{3} - x + arctan (x)) + 6 arctan (x) : - 2 x^{3} + 6 x

= - 2 x^{3} + 6 x

Füge eine Konstante zur Lösung hinzu

= - 2 x^{3} + 6 x + C

\int \frac{1}{e ^{10 x}} d x

\frac{d}{d x} (3 x^{- 3})

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

s l o p e in t erce pt (- 2, 7), (0, 5)

\frac{d}{d v} (\frac{u}{v})