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

Lösung

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

2 (2 1 + x + \frac{2}{1 + x}) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

Multipliziere aus

\frac{v - 1}{v ^{\frac{3}{2}}} : \frac{1}{v ^{\frac{1}{2}}} - \frac{1}{v ^{\frac{3}{2}}}

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

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

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

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

= 2 (2 v^{\frac{1}{2}} - (- \frac{2}{v ^{\frac{1}{2}}}))

Ersetze zurück

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

Vereinfache

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

= 2 (2 1 + x + \frac{2}{1 + x})

Füge eine Konstante zur Lösung hinzu

= 2 (2 1 + x + \frac{2}{1 + x}) + C

\frac{d ^{3} y}{d x ^{3}} = 4 x^{3} + x

\frac{d ^{2}}{d x ^{2}} (\frac{6}{7 x ^{4}})

\int 2 π x^{3} 1 + 9 x^{4} d x

\frac{\partial}{\partial l} (9 l^{\frac{2}{3}} k^{\frac{1}{3}})

y^{^{''}} + 9 y = 24 sin (3 t)