integral of 8x^5-5x^4-4x^3-6x^2-2x-3

Lösung

\int 8 x^{5} - 5 x^{4} - 4 x^{3} - 6 x^{2} - 2 x - 3 d x

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

Schritte zur Lösung

\int 8 x^{5} - 5 x^{4} - 4 x^{3} - 6 x^{2} - 2 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 8 x^{5} d x - \int 5 x^{4} d x - \int 4 x^{3} d x - \int 6 x^{2} d x - \int 2 x d x - \int 3 d x

\int 8 x^{5} d x = \frac{4 x ^{6}}{3}

\int 5 x^{4} d x = x^{5}

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

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

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

\int 3 d x = 3 x

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

Füge eine Konstante zur Lösung hinzu

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

\int (6 x^{2} + 1) ln (5 x^{3}) d x

\int - \frac{e ^{x}}{x} d x

\int (e^{x}) \cdot sin (n x) d x

\int 3 t (t + 1)^{3} d t

\int \frac{cos ( ln ( 4 x ) )}{x} d x