integral of 4-6x-40x^3

Lösung

\int 4 - 6 x - 40 x^{3} d x

4 x - 3 x^{2} - 10 x^{4} + C

Schritte zur Lösung

\int 4 - 6 x - 40 x^{3} 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 4 d x - \int 6 x d x - \int 40 x^{3} d x

\int 4 d x = 4 x

\int 6 x d x = 3 x^{2}

\int 40 x^{3} d x = 10 x^{4}

= 4 x - 3 x^{2} - 10 x^{4}

Füge eine Konstante zur Lösung hinzu

= 4 x - 3 x^{2} - 10 x^{4} + C

\int_{0}^{2} 1 + e^{- 2 x} d x

n = 1 \sum \infty \frac{3}{( - 2 ) ^{n}}

\int_{0}^{\infty} \frac{1}{5 + x} d x

d er i v a t i v e f (x) = x^{2}

d er i v a t i v e 5 arcsin (\frac{x - 1}{x + 1})