integral of (x^5+8x^3-9x^2+7x-5)

Lösung

\int (x^{5} + 8 x^{3} - 9 x^{2} + 7 x - 5) d x

\frac{x ^{6}}{6} + 2 x^{4} - 3 x^{3} + \frac{7 x ^{2}}{2} - 5 x + C

Schritte zur Lösung

\int x^{5} + 8 x^{3} - 9 x^{2} + 7 x - 5 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 x^{5} d x + \int 8 x^{3} d x - \int 9 x^{2} d x + \int 7 x d x - \int 5 d x

\int x^{5} d x = \frac{x ^{6}}{6}

\int 8 x^{3} d x = 2 x^{4}

\int 9 x^{2} d x = 3 x^{3}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 5 d x = 5 x

= \frac{x ^{6}}{6} + 2 x^{4} - 3 x^{3} + \frac{7 x ^{2}}{2} - 5 x

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{6}}{6} + 2 x^{4} - 3 x^{3} + \frac{7 x ^{2}}{2} - 5 x + C

\int (e^{x} + \frac{1}{e ^{x}})^{2} d x

\int_{0}^{1} cos (x^{2}) d x

2 t \cdot \frac{d y}{d t} - 4 y = t

\int (\frac{x ^{2}}{4} + 4 x + 19)^{2} d x

(8 y^{2} - x y) d x + x^{2} d y = 0