integral of (9-9x)/(1-sqrt(x))

Lösung

\int \frac{9 - 9 x}{1 - x} d x

18 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

Schritte zur Lösung

\int \frac{9 - 9 x}{1 - x} d x

Wende U-Substitution an

= \int 18 u (u + 1) d u

Entferne die Konstante:

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

= 18 \cdot \int u (u + 1) d u

Multipliziere aus

u (u + 1) : u^{2} + u

= 18 \cdot \int u^{2} + 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

= 18 (\int u^{2} d u + \int u d u)

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

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

= 18 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Setze in

u = x

ein

= 18 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Vereinfache

18 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 18 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 18 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= 18 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

t an g e n t y = x, at x = 4

\frac{d y}{d x} = \frac{x}{9 y}

d er i v a t i v e \frac{3 x ^{\frac{3}{5}} - x}{x - 1}

\int \frac{x + 1}{x ^{2} + 2 x + 5} d x

\int - 5 x^{4} d x