integral of ((sqrt(3x)+4))/(sqrt(3x)-2)

Lösung

\int \frac{( 3 x + 4 )}{3 x - 2} d x

\frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28) + C

Schritte zur Lösung

\int \frac{3 x + 4}{3 x - 2} d x

Wende U-Substitution an

= \int \frac{u + 4}{3 ( u - 2 )} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{u + 4}{u - 2} d u

Multipliziere aus

\frac{u + 4}{u - 2} : \frac{u}{u - 2} + \frac{4}{u - 2}

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}{3} (\int \frac{u}{u - 2} d u + \int \frac{4}{u - 2} d u)

\int \frac{u}{u - 2} d u = u + 4 u + 8 ln u - 2 - 12

\int \frac{4}{u - 2} d u = 8 (u - 2 + 2 ln u - 2)

= \frac{1}{3} (u + 4 u + 8 ln u - 2 - 12 + 8 (u - 2 + 2 ln u - 2))

Setze in

u = 3 x

ein

= \frac{1}{3} (3 x + 4 3 x + 8 ln 3 x - 2 - 12 + 8 (3 x - 2 + 2 ln 3 x - 2))

Vereinfache

\frac{1}{3} (3 x + 4 3 x + 8 ln 3 x - 2 - 12 + 8 (3 x - 2 + 2 ln 3 x - 2)) : \frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28)

= \frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28) + C

\int x^{3} cos (y) d y

\int e^{i w x} d x

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

\int 4 x^{2} d x

\int x^{3} + x^{2} d x