integral of (t^2-1)(4+3t)

Lösung

\int (t^{2} - 1) (4 + 3 t) d t

\frac{4 t ^{3}}{3} + \frac{3 t ^{4}}{4} - 4 t - \frac{3 t ^{2}}{2} + C

Schritte zur Lösung

\int (t^{2} - 1) (4 + 3 t) d t

Multipliziere aus

(t^{2} - 1) (4 + 3 t) : 4 t^{2} + 3 t^{3} - 4 - 3 t

= \int 4 t^{2} + 3 t^{3} - 4 - 3 t d t

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 t^{2} d t + \int 3 t^{3} d t - \int 4 d t - \int 3 t d t

\int 4 t^{2} d t = \frac{4 t ^{3}}{3}

\int 3 t^{3} d t = \frac{3 t ^{4}}{4}

\int 4 d t = 4 t

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

= \frac{4 t ^{3}}{3} + \frac{3 t ^{4}}{4} - 4 t - \frac{3 t ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{4 t ^{3}}{3} + \frac{3 t ^{4}}{4} - 4 t - \frac{3 t ^{2}}{2} + C

t an g e n t f (x) = x^{4}, at x = 1

t an g e n t y = \frac{6}{x}, (25, \frac{6}{5})

x \to \infty lim (\frac{( 2 x ^{2} + 1 )}{3 x - 5})

l a pl a ce t r an s f or m cos (t + \frac{π}{4})

x^{2} \frac{d y}{d x} - 2 x y = 4 y^{4}, y (1) = \frac{1}{3}