integral of (x^2+1)/(\sqrt[3]{x)}

Lösung

\int \frac{x ^{2} + 1}{3 x} d x

\frac{3}{8} x^{\frac{8}{3}} + \frac{3}{2} x^{\frac{2}{3}} + C

Schritte zur Lösung

\int \frac{x ^{2} + 1}{3 x} d x

Multipliziere aus

\frac{x ^{2} + 1}{3 x} : x^{\frac{5}{3}} + \frac{1}{3 x}

= \int x^{\frac{5}{3}} + \frac{1}{3 x} 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^{\frac{5}{3}} d x + \int \frac{1}{3 x} d x

\int x^{\frac{5}{3}} d x = \frac{3}{8} x^{\frac{8}{3}}

\int \frac{1}{3 x} d x = \frac{3}{2} x^{\frac{2}{3}}

= \frac{3}{8} x^{\frac{8}{3}} + \frac{3}{2} x^{\frac{2}{3}}

Füge eine Konstante zur Lösung hinzu

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

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

\int \frac{e ^{\frac{4}{x}}}{x ^{2}} d x

\int \frac{1}{x ^{2}} - 3 x d x

x \to - \infty lim (xx)

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