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

Lösung

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

\frac{3}{8} (2 + x)^{\frac{8}{3}} - \frac{12}{5} (2 + x)^{\frac{5}{3}} + 6 (2 + x)^{\frac{2}{3}} + \frac{6}{7} (2 + x)^{\frac{7}{6}} + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

\frac{( u - 2 ) ^{2} + u}{3 u} : u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u} + 6 u

= \int u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u} + 6 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

= \int u^{\frac{5}{3}} d u - \int 4 u^{\frac{2}{3}} d u + \int \frac{4}{3 u} d u + \int 6 u d u

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

\int 4 u^{\frac{2}{3}} d u = \frac{12}{5} u^{\frac{5}{3}}

\int \frac{4}{3 u} d u = 6 u^{\frac{2}{3}}

\int 6 u d u = \frac{6}{7} u^{\frac{7}{6}}

= \frac{3}{8} u^{\frac{8}{3}} - \frac{12}{5} u^{\frac{5}{3}} + 6 u^{\frac{2}{3}} + \frac{6}{7} u^{\frac{7}{6}}

Setze in

u = 2 + x

ein

= \frac{3}{8} (2 + x)^{\frac{8}{3}} - \frac{12}{5} (2 + x)^{\frac{5}{3}} + 6 (2 + x)^{\frac{2}{3}} + \frac{6}{7} (2 + x)^{\frac{7}{6}}

Füge eine Konstante zur Lösung hinzu

= \frac{3}{8} (2 + x)^{\frac{8}{3}} - \frac{12}{5} (2 + x)^{\frac{5}{3}} + 6 (2 + x)^{\frac{2}{3}} + \frac{6}{7} (2 + x)^{\frac{7}{6}} + C

d er i v a t i v e x^{2} \cdot 2

x \to \infty lim (- 2 e^{- x} + 2)

\int 3 x y d y

im pl i c i t \frac{d y}{d x}, y = (sin (x))^{6 x}

\frac{\partial}{\partial x} (x^{2} + y^{2} - 9)