integral of (-4x^6-5x^5-2x^2+5-e^x)

Lösung

\int (- 4 x^{6} - 5 x^{5} - 2 x^{2} + 5 - e^{x}) d x

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

Schritte zur Lösung

\int - 4 x^{6} - 5 x^{5} - 2 x^{2} + 5 - e^{x} 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 x^{6} d x - \int 5 x^{5} d x - \int 2 x^{2} d x + \int 5 d x - \int e^{x} d x

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

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

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

\int 5 d x = 5 x

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

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

Füge eine Konstante zur Lösung hinzu

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

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 17, y^{^{'}} (0) = \frac{55}{4}

\frac{\partial}{\partial y} (ln (1 + x^{2} + e^{y} + z))

in v erse l a pl a ce \frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}

y^{^{'}} = t y - 2 t + y - 2

l a pl a ce t r an s f or m 3 t^{3} - 2 t^{2} - 7 t + 2