integral x/((2+2x))

Lösung

integral

\frac{x}{( 2 + 2 x )}

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

Schritte zur Lösung

\int \frac{x}{2 + 2 x} d x

Wende U-Substitution an

= \int \frac{u - 2}{4 u} d u

Entferne die Konstante:

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

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

Multipliziere aus

\frac{u - 2}{u} : 1 - \frac{2}{u}

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

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{1}{4} (u - 2 ln ∣ u ∣)

Setze in

u = 2 + 2 x

ein

= \frac{1}{4} (2 + 2 x - 2 ln ∣ 2 + 2 x ∣)

Vereinfache

\frac{1}{4} (2 + 2 x - 2 ln ∣ 2 + 2 x ∣) : \frac{1}{2} (1 + x - ln (2 ∣ x + 1 ∣))

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{1}{e ^{x} + 1}

in t e g r a l x + 16

in t e g r a l \frac{1}{( x ^{2} - 4 x )}

in t e g r a l (2 x + 3)^{4}

in t e g r a l e \cdot x^{2}