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

Lösung

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

\frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

Multipliziere aus

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

= 3 \cdot \int v^{\frac{3}{2}} - 2 v + \frac{1}{v} d v

Wende die Summenregel an:

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

= 3 (\int v^{\frac{3}{2}} d v - \int 2 v d v + \int \frac{1}{v} d v)

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

\int 2 v d v = \frac{4}{3} v^{\frac{3}{2}}

\int \frac{1}{v} d v = 2 v

= 3 (\frac{2}{5} v^{\frac{5}{2}} - \frac{4}{3} v^{\frac{3}{2}} + 2 v)

Ersetze zurück

= 3 (\frac{2}{5} (1 + 3 x)^{\frac{5}{2}} - \frac{4}{3} (1 + 3 x)^{\frac{3}{2}} + 2 1 + 3 x)

Vereinfache

3 (\frac{2}{5} (1 + 3 x)^{\frac{5}{2}} - \frac{4}{3} (1 + 3 x)^{\frac{3}{2}} + 2 1 + 3 x) : \frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x

= \frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x

Füge eine Konstante zur Lösung hinzu

= \frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x + C

x \to 1 lim (\frac{7}{x ^{3} - 1})

x \to 0 lim (\frac{1}{x} - \frac{1}{x ^{2} - x})

\frac{\partial}{\partial z} (5 x sin (y - z))

a re a f (x) = \frac{1}{7 ( 1 + x ^{2} )}, g (x) = \frac{1}{14 x ^{2}}

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