integral of ((9+x+x^2))/(sqrt(x))

Lösung

\int \frac{( 9 + x + x ^{2} )}{x} d x

18 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

Schritte zur Lösung

\int \frac{9 + x + x ^{2}}{x} d x

Multipliziere aus

\frac{9 + x + x ^{2}}{x} : \frac{9}{x} + x + x^{\frac{3}{2}}

= \int \frac{9}{x} + x + x^{\frac{3}{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 \frac{9}{x} d x + \int x d x + \int x^{\frac{3}{2}} d x

\int \frac{9}{x} d x = 18 x

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

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

= 18 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}}

Füge eine Konstante zur Lösung hinzu

= 18 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

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

\int t^{5} - 7 t d t

\frac{\partial}{\partial x} (2 x^{2} + 4 y^{2} + 1)

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tan (t) y^{^{'}} - y = 1