integral of (2x^2+1/((4x)^2))

Lösung

\int (2 x^{2} + \frac{1}{( 4 x ) ^{2}}) d x

\frac{2 x ^{3}}{3} + \frac{1}{16 x} - \frac{1}{8 x} + C

Schritte zur Lösung

\int 2 x^{2} + \frac{1}{( 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^{2} d x + \int \frac{1}{( 4 x ) ^{2}} d x

\int 2 x^{2} d x = \frac{2 x ^{3}}{3}

\int \frac{1}{( 4 x ) ^{2}} d x = \frac{1}{16 x} - \frac{1}{8 x}

= \frac{2 x ^{3}}{3} + \frac{1}{16 x} - \frac{1}{8 x}

Füge eine Konstante zur Lösung hinzu

= \frac{2 x ^{3}}{3} + \frac{1}{16 x} - \frac{1}{8 x} + C

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

\frac{\partial}{\partial z} (x cos (y))

x \to 0 + lim ((1 - x)^{(\frac{1}{x})})

2 x y^{^{'}} - 6 y = x^{- 4}

x \to (\frac{π}{2}) - lim (sec (x))