integral of 1/(6+2sqrt(x))

Lösung

\int \frac{1}{6 + 2 x} d x

3 + x - 3 ln 3 + x + C

Schritte zur Lösung

\int \frac{1}{6 + 2 x} d x

Wende U-Substitution an

= \int \frac{u}{3 + u} d u

Wende U-Substitution an

= \int \frac{v - 3}{v} d v

Multipliziere aus

\frac{v - 3}{v} : 1 - \frac{3}{v}

= \int 1 - \frac{3}{v} d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d v - \int \frac{3}{v} d v

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= v - 3 ln ∣ v ∣

Ersetze zurück

= 3 + x - 3 ln 3 + x

Füge eine Konstante zur Lösung hinzu

= 3 + x - 3 ln 3 + x + C

\frac{d}{d x} ((x^{5} + 2)^{2} (x^{3} + 4)^{4})

n = 4 \sum \infty 3 (- \frac{1}{2})^{n}

x \to 0 lim (x^{\frac{1}{3}})

t an g e n t (x^{2} + y^{2})^{2} = 4 x^{2} y, (1, 1)

\frac{\partial}{\partial x} (x^{3} + 2^{y} + x^{y})