integral of (3x+5)/(sqrt(x(2x-1)))

Lösung

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

\frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1) + C

Schritte zur Lösung

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

Wende U-Substitution an

= \int \frac{3 u + 13}{2 2 u ( u + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{2 2} \cdot \int \frac{3 u + 13}{u ( u + 1 )} d u

Multipliziere aus

\frac{3 u + 13}{u ( u + 1 )} : \frac{3 u}{u ^{2} + u} + \frac{13}{u ^{2} + u}

= \frac{1}{2 2} \cdot \int \frac{3 u}{u ^{2} + u} + \frac{13}{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

= \frac{1}{2 2} (\int \frac{3 u}{u ^{2} + u} d u + \int \frac{13}{u ^{2} + u} d u)

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

\int \frac{13}{u ^{2} + u} d u = 26 ln u + 1 + u

= \frac{1}{2 2} (3 (\frac{u}{u + 1} (u + 1) - ln u + 1 + u) + 26 ln u + 1 + u)

Setze in

u = 2 x - 1

ein

= \frac{1}{2 2} (3 (\frac{2 x - 1}{2 x - 1 + 1} (2 x - 1 + 1) - ln 2 x - 1 + 1 + 2 x - 1) + 26 ln 2 x - 1 + 1 + 2 x - 1)

Vereinfache

\frac{1}{2 2} (3 (\frac{2 x - 1}{2 x - 1 + 1} (2 x - 1 + 1) - ln 2 x - 1 + 1 + 2 x - 1) + 26 ln 2 x - 1 + 1 + 2 x - 1) : \frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1)

= \frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1) + C

d er i v a t i v e t^{3} e^{t} t

n \to \infty lim (c^{n})

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

(3 cos (t))^{^{'}}

\frac{d}{d x} (x^{3} - 9 x + 3)