integral of (x^4-2x^2+3)/(x^2)

Lösung

\int \frac{x ^{4} - 2 x ^{2} + 3}{x ^{2}} d x

\frac{x ^{3}}{3} - 2 x - \frac{3}{x} + C

Schritte zur Lösung

\int \frac{x ^{4} - 2 x ^{2} + 3}{x ^{2}} d x

Multipliziere aus

\frac{x ^{4} - 2 x ^{2} + 3}{x ^{2}} : x^{2} - 2 + \frac{3}{x ^{2}}

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

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

\int 2 d x = 2 x

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

= \frac{x ^{3}}{3} - 2 x - \frac{3}{x}

Füge eine Konstante zur Lösung hinzu

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

n = 1 \sum \infty \frac{n ^{n}}{n !}

4 \frac{d y}{d x} + 28 y = 7

(x^{l n (x)})^{^{'}}

in v erse l a pl a ce \frac{( s + 3 )}{( s ^{2} )}

\int_{0}^{\frac{π}{3}} x tan^{2} (x) d x