integral of xsqrt(1+2x)

Lösung

\int x 1 + 2 x d x

\frac{1}{15} (3 x - 1) (2 x + 1)^{\frac{3}{2}} + C

Schritte zur Lösung

\int x 1 + 2 x d x

Wende U-Substitution an

= \int \frac{u ( u - 1 )}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int u (u - 1) d u

Multipliziere aus

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

= \frac{1}{4} \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

= \frac{1}{4} (\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}}

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

Setze in

u = 1 + 2 x

ein

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

Vereinfache

\frac{1}{4} (\frac{2}{5} (1 + 2 x)^{\frac{5}{2}} - \frac{2}{3} (1 + 2 x)^{\frac{3}{2}}) : \frac{1}{15} (3 x - 1) (2 x + 1)^{\frac{3}{2}}

= \frac{1}{15} (3 x - 1) (2 x + 1)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{15} (3 x - 1) (2 x + 1)^{\frac{3}{2}} + C

\int_{0}^{3} \frac{1}{5 x + 1} d x

\int \frac{e ^{x} - e ^{- x}}{e ^{x} - e ^{- x}} d x

d er i v a t i v e y = ∣ x ∣

t an g e n t 7 x + 1, at x = 9

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