integral of sqrt(1+(16)/(9*x^{2/3))}

Lösung

\int 1 + \frac{16}{9 \cdot x ^{\frac{2}{3}}} d x

\frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

Schritte zur Lösung

\int 1 + \frac{16}{9 x ^{\frac{2}{3}}} d x

Vereinfache

1 + \frac{16}{9 x ^{\frac{2}{3}}} : \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}}

= \int \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Wende U-Substitution an

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{2}}}{6} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \frac{1}{6} \cdot \int u^{\frac{1}{2}} d u

Wende die Potenzregel an

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Setze in

u = 9 x^{\frac{2}{3}} + 16

ein

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

Vereinfache

\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} : \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

\int_{0}^{7} 1 + 9 x d x

6 y^{^{'''}} + 3 y^{^{''}} = 0

t an g e n t \frac{3}{( x + 6 )}, at x = - 4

y^{^{'}} = 7 x + y

\frac{d y}{d x} + 3 x y^{2} = 0, y (0) = 1