integral of 2x+4x^{-2}

Lösung

\int 2 x + 4 x^{- 2} d x

x^{2} - \frac{4}{x} + C

Schritte zur Lösung

\int 2 x + 4 x^{- 2} d x

4 x^{- 2} = \frac{4}{x ^{2}}

= \int 2 x + \frac{4}{x ^{2}} 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 2 x d x + \int \frac{4}{x ^{2}} d x

\int 2 x d x = x^{2}

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

= x^{2} - \frac{4}{x}

Füge eine Konstante zur Lösung hinzu

= x^{2} - \frac{4}{x} + C

x \to 6 lim (9 x + ∣ x - 6 ∣)

d er i v a t i v e y = 3^{x}

\frac{d y}{d x} = \frac{x y + y + x + 1}{x y + 2 x - y - 2}

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

in v erse l a pl a ce \frac{( 2 s + 1 )}{s ^{2} + 2 s + 5}