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

Lösung

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

\frac{16 t ^{5}}{5} - 8 t^{3} + 9 t + C

Schritte zur Lösung

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

Multipliziere aus

(4 t^{2} - 3)^{2} : 16 t^{4} - 24 t^{2} + 9

= \int 16 t^{4} - 24 t^{2} + 9 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 16 t^{4} d t - \int 24 t^{2} d t + \int 9 d t

\int 16 t^{4} d t = \frac{16 t ^{5}}{5}

\int 24 t^{2} d t = 8 t^{3}

\int 9 d t = 9 t

= \frac{16 t ^{5}}{5} - 8 t^{3} + 9 t

Füge eine Konstante zur Lösung hinzu

= \frac{16 t ^{5}}{5} - 8 t^{3} + 9 t + C

a re a f (x) = x^{2}, [1, 5]

d er i v a t i v e \frac{8 x ^{2} + 2}{x ^{2}}

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