integral of (9x^2)/2-5cos(x)+7

Lösung

\int \frac{9 x ^{2}}{2} - 5 cos (x) + 7 d x

\frac{3 x ^{3}}{2} - 5 sin (x) + 7 x + C

Schritte zur Lösung

\int \frac{9 x ^{2}}{2} - 5 cos (x) + 7 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 \frac{9 x ^{2}}{2} d x - \int 5 cos (x) d x + \int 7 d x

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

\int 5 cos (x) d x = 5 sin (x)

\int 7 d x = 7 x

= \frac{3 x ^{3}}{2} - 5 sin (x) + 7 x

Füge eine Konstante zur Lösung hinzu

= \frac{3 x ^{3}}{2} - 5 sin (x) + 7 x + C

in v erse l a pl a ce - \frac{1}{s ^{2}}

\frac{\partial}{\partial x} (y + \frac{1}{x + y})

\frac{d}{d x} (lo g_{3} (sec (2 x)))

x \to 0 lim (tan (x) - sec (x))

l a pl a ce t r an s f or m (t e)^{3} e^{- 2 t} e^{- 2}