integral from 2 to 4 of (6+u^2)/(u^3)

Lösung

\int_{2}^{4} \frac{6 + u ^{2}}{u ^{3}} d u

\frac{9}{16} + ln (2)

Dezimale

1.25564 \dots

Schritte zur Lösung

\int_{2}^{4} \frac{6 + u ^{2}}{u ^{3}} d u

Multipliziere aus

\frac{6 + u ^{2}}{u ^{3}} : \frac{6}{u ^{3}} + \frac{1}{u}

= \int_{2}^{4} \frac{6}{u ^{3}} + \frac{1}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{4} \frac{6}{u ^{3}} d u + \int_{2}^{4} \frac{1}{u} d u

\int_{2}^{4} \frac{6}{u ^{3}} d u = \frac{9}{16}

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= \frac{9}{16} + ln (2)

x \to 4 lim (2 x + 1)

\frac{d}{d y} (e^{y^{2}})

a re a 4 - x^{2}, 0, 2

x \to \frac{π}{2} lim (sin (2 x))

\int \frac{2 x - 5}{x ^{2} + 9} d x