integral of 1/(\sqrt[4]{5-x)+sqrt(5-x)}

Lösung

\int \frac{1}{4 5 - x + 5 - x} d x

- 4 (\frac{( 1 + 4 5 - x ) ^{2}}{2} - 2 (1 + 4 5 - x) + ln 1 + 4 5 - x) + C

Schritte zur Lösung

\int \frac{1}{4 5 - x + 5 - x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= - \int \frac{4 v ^{2}}{1 + v} d v

Entferne die Konstante:

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

= - 4 \cdot \int \frac{v ^{2}}{1 + v} d v

Wende U-Substitution an

= - 4 \cdot \int \frac{( w - 1 ) ^{2}}{w} d w

Multipliziere aus

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

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

Wende die Summenregel an:

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

= - 4 (\int w d w - \int 2 d w + \int \frac{1}{w} d w)

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

\int 2 d w = 2 w

\int \frac{1}{w} d w = ln ∣ w ∣

= - 4 (\frac{w ^{2}}{2} - 2 w + ln ∣ w ∣)

Ersetze zurück

= - 4 (\frac{( 1 + 4 5 - x ) ^{2}}{2} - 2 (1 + 4 5 - x) + ln 1 + 4 5 - x)

Füge eine Konstante zur Lösung hinzu

= - 4 (\frac{( 1 + 4 5 - x ) ^{2}}{2} - 2 (1 + 4 5 - x) + ln 1 + 4 5 - x) + C

\frac{d}{d θ} (\frac{5}{sin ( θ ) - 2 cos ( θ )})

x^{2} \frac{d y}{d x} = x - x y

\int x^{2} - 6 x d x

d er i v a t i v e - \frac{1}{3} x^{2}

\frac{d}{d x} (ln (9 x^{2} - 11))