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

Lösung

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

\frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\frac{2}{3}}) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{( u - 1 ) ^{2}}{8 3 u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{( u - 1 ) ^{2}}{3 u} d u

Multipliziere aus

\frac{( u - 1 ) ^{2}}{3 u} : u^{\frac{5}{3}} - 2 u^{\frac{2}{3}} + \frac{1}{3 u}

= \frac{1}{8} \cdot \int u^{\frac{5}{3}} - 2 u^{\frac{2}{3}} + \frac{1}{3 u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int u^{\frac{5}{3}} d u - \int 2 u^{\frac{2}{3}} d u + \int \frac{1}{3 u} d u)

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

\int 2 u^{\frac{2}{3}} d u = \frac{6}{5} u^{\frac{5}{3}}

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

= \frac{1}{8} (\frac{3}{8} u^{\frac{8}{3}} - \frac{6}{5} u^{\frac{5}{3}} + \frac{3}{2} u^{\frac{2}{3}})

Setze in

u = 1 + 2 x

ein

= \frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\frac{2}{3}})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\frac{2}{3}}) + C

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

2 (x y - y) d x - x d y = 0

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\int \frac{π}{2} - 5 x^{- 0.5} d x