integral of (x+1)/(4x)

Lösung

\int \frac{x + 1}{4 x} d x

\frac{1}{4} (x + ln ∣ x ∣) + C

Schritte zur Lösung

\int \frac{x + 1}{4 x} d x

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int \frac{x + 1}{x} d x

Multipliziere aus

\frac{x + 1}{x} : 1 + \frac{1}{x}

= \frac{1}{4} \cdot \int 1 + \frac{1}{x} d x

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 1 d x + \int \frac{1}{x} d x)

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{4} (x + ln ∣ x ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (x + ln ∣ x ∣) + C

d er i v a t i v e f (x) = (1 + 5 x^{2}) (x - x^{2})

\int \frac{e ^{- 1 + x + l n (1 - x)}}{x ^{2}} d x

y^{^{'}} = - 6 x y

y^{^{'}} (x) = \frac{9 x ^{2}}{9 y ^{2} - 11}, y (1) = 0

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