integral of 1/(1+\sqrt[3]{x)}

Lösung

\int \frac{1}{1 + 3 x} d x

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

Schritte zur Lösung

\int \frac{1}{1 + 3 x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 3 \cdot \int \frac{( v - 1 ) ^{2}}{v} d v

Multipliziere aus

\frac{( v - 1 ) ^{2}}{v} : v - 2 + \frac{1}{v}

= 3 \cdot \int v - 2 + \frac{1}{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

= 3 (\int v d v - \int 2 d v + \int \frac{1}{v} d v)

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 3 (\frac{v ^{2}}{2} - 2 v + ln ∣ v ∣)

Ersetze zurück

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t \frac{5}{4 + x ^{2}}, at x = 2

\int 3 csc^{2} (x) d x

y^{^{'}} + cot (x) y - x e^{x} = 0

\int (2 x + 1) e^{x (x + 1)} d x

d er i v a t i v e (10 x^{2} - 20 x + 20) e^{x}