integral of ((4u^2-8))/(2u^2+1)

Lösung

\int \frac{( 4 u ^{2} - 8 )}{2 u ^{2} + 1} d u

2 u - 52 arctan (2 u) + C

Schritte zur Lösung

\int \frac{4 u ^{2} - 8}{2 u ^{2} + 1} d u

Multipliziere aus

\frac{4 u ^{2} - 8}{2 u ^{2} + 1} : \frac{4 u ^{2}}{2 u ^{2} + 1} - \frac{8}{2 u ^{2} + 1}

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{4 u ^{2}}{2 u ^{2} + 1} d u - \int \frac{8}{2 u ^{2} + 1} d u

\int \frac{4 u ^{2}}{2 u ^{2} + 1} d u = 2 (- arctan (2 u) + 2 u)

\int \frac{8}{2 u ^{2} + 1} d u = 42 arctan (2 u)

= 2 (- arctan (2 u) + 2 u) - 42 arctan (2 u)

Vereinfache

2 (- arctan (2 u) + 2 u) - 42 arctan (2 u) : 2 u - 52 arctan (2 u)

= 2 u - 52 arctan (2 u)

Füge eine Konstante zur Lösung hinzu

= 2 u - 52 arctan (2 u) + C

\frac{d}{d x} (\frac{x ^{2}}{4 - x ^{2}})

d er i v a t i v e - \frac{24}{x ^{4}}

\int p (p + 3)^{6} d p

x \to 4 lim (x - 4)

x \to 5 lim (25 - x^{2})