integral of 2xsqrt(x-1)

Lösung

\int 2 x x - 1 d x

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

Schritte zur Lösung

\int 2 x x - 1 d x

Entferne die Konstante:

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

= 2 \cdot \int x x - 1 d x

Wende U-Substitution an

= 2 \cdot \int (u + 1) u d u

Multipliziere aus

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

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

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

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

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

= 2 (\frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Setze in

u = x - 1

ein

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (x (3 - 2 x)^{5})

y^{^{''}} + 4 y = t^{2} e^{3 t} + 3

x y^{^{'}} = y^{2} + 1

in v erse l a pl a ce \frac{3}{s + 2}

(\frac{x}{1 + x ^{2}})^{^{'}}