integral from 1 to 3 of (x^2+1)^3

Lösung

\int_{1}^{3} (x^{2} + 1)^{3} d x

\frac{16992}{35}

Dezimale

485.48571 \dots

Schritte zur Lösung

\int_{1}^{3} (x^{2} + 1)^{3} d x

Multipliziere aus

(x^{2} + 1)^{3} : x^{6} + 3 x^{4} + 3 x^{2} + 1

= \int_{1}^{3} x^{6} + 3 x^{4} + 3 x^{2} + 1 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_{1}^{3} x^{6} d x + \int_{1}^{3} 3 x^{4} d x + \int_{1}^{3} 3 x^{2} d x + \int_{1}^{3} 1 d x

\int_{1}^{3} x^{6} d x = \frac{2186}{7}

\int_{1}^{3} 3 x^{4} d x = \frac{726}{5}

\int_{1}^{3} 3 x^{2} d x = 26

\int_{1}^{3} 1 d x = 2

= \frac{2186}{7} + \frac{726}{5} + 26 + 2

Vereinfache

\frac{2186}{7} + \frac{726}{5} + 26 + 2 : \frac{16992}{35}

= \frac{16992}{35}

\frac{d ^{2}}{d x ^{2}} (8 x cos (x^{2}))

\frac{d}{d x} (sec^{2} (x) - 3)

t an g e n t - 3 x^{2}, at x = 8

d er i v a t i v e f (t) = \frac{3 t}{t - 6}

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2}})