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

Lösung

\int \frac{x ^{2} + 1}{x ^{2}} d x

- \frac{1}{x} + x + C

Schritte zur Lösung

\int \frac{x ^{2} + 1}{x ^{2}} d x

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

= \int \frac{1}{x ^{2}} + 1 d x

Wende Exponentenregel an:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2}} = x^{- 2}

= \int 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 x^{- 2} d x + \int 1 d x

\int x^{- 2} d x = - \frac{1}{x}

\int 1 d x = x

= - \frac{1}{x} + x

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{x} + x + C

x e^{y} d y + \frac{x ^{2} + 1}{y} d x = 0

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

\frac{d}{d x} (x^{2} sin (x) - x cos (x))

d er i v a t i v e (3 x^{3} + 2 x) (x - 2) (x + 4)

cos (x) d x - sin (y) d y = 0