integral of 1/(\sqrt[8]{x^5)-\sqrt[8]{x}}

Lösung

\int \frac{1}{8 x ^{5} - 8 x} d x

8 (\frac{1}{3} x^{\frac{3}{8}} + \frac{1}{2} arctan (8 x) - \frac{1}{4} ln 8 x + 1 + \frac{1}{4} ln 8 x - 1) + C

Schritte zur Lösung

\int \frac{1}{8 x ^{5} - 8 x} d x

Wende U-Substitution an

= \int \frac{8 u ^{6}}{u ^{4} - 1} d u

Entferne die Konstante:

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

= 8 \cdot \int \frac{u ^{6}}{u ^{4} - 1} d u

Lange Division

\frac{u ^{6}}{u ^{4} - 1} : u^{2} + \frac{u ^{2}}{u ^{4} - 1}

= 8 \cdot \int u^{2} + \frac{u ^{2}}{u ^{4} - 1} 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

= 8 (\int u^{2} d u + \int \frac{u ^{2}}{u ^{4} - 1} d u)

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

\int \frac{u ^{2}}{u ^{4} - 1} d u = \frac{1}{2} arctan (u) - \frac{1}{4} ln ∣ u + 1 ∣ + \frac{1}{4} ln ∣ u - 1 ∣

= 8 (\frac{u ^{3}}{3} + \frac{1}{2} arctan (u) - \frac{1}{4} ln ∣ u + 1 ∣ + \frac{1}{4} ln ∣ u - 1 ∣)

Setze in

u = 8 x

ein

= 8 (\frac{( 8 x ) ^{3}}{3} + \frac{1}{2} arctan (8 x) - \frac{1}{4} ln 8 x + 1 + \frac{1}{4} ln 8 x - 1)

Vereinfache

8 (\frac{( 8 x ) ^{3}}{3} + \frac{1}{2} arctan (8 x) - \frac{1}{4} ln 8 x + 1 + \frac{1}{4} ln 8 x - 1) : 8 (\frac{1}{3} x^{\frac{3}{8}} + \frac{1}{2} arctan (8 x) - \frac{1}{4} ln 8 x + 1 + \frac{1}{4} ln 8 x - 1)

= 8 (\frac{1}{3} x^{\frac{3}{8}} + \frac{1}{2} arctan (8 x) - \frac{1}{4} ln 8 x + 1 + \frac{1}{4} ln 8 x - 1)

Füge eine Konstante zur Lösung hinzu

= 8 (\frac{1}{3} x^{\frac{3}{8}} + \frac{1}{2} arctan (8 x) - \frac{1}{4} ln 8 x + 1 + \frac{1}{4} ln 8 x - 1) + C

d x + e^{4 x} d y = 0

d er i v a t i v e y = sin (x^{5})

d er i v a t i v e x e^{a x}

\frac{d}{d x} (ln (tanh (2 x)))

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