integral of ((x-2)^4)/(sqrt(4x-x^2))

Lösung

\int \frac{( x - 2 ) ^{4}}{4 x - x ^{2}} d x

- \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1)) + C

Schritte zur Lösung

\int \frac{( x - 2 ) ^{4}}{4 x - x ^{2}} d x

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= \int 16 sin^{4} (v) d v

Entferne die Konstante:

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

= 16 \cdot \int sin^{4} (v) d v

Wende integrale Reduktion an

= 16 (- \frac{cos ( v ) sin ^{3} ( v )}{4} + \frac{3}{4} \cdot \int sin^{2} (v) d v)

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

= 16 (- \frac{cos ( v ) sin ^{3} ( v )}{4} + \frac{3}{4} \cdot \frac{1}{2} (v - \frac{1}{2} sin (2 v)))

Ersetze zurück

= 16 (- \frac{cos ( arcsin ( \frac{1}{2} ( x - 2 ) ) ) sin ^{3} ( arcsin ( \frac{1}{2} ( x - 2 ) ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (\frac{1}{2} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 2)))))

Vereinfache

16 (- \frac{cos ( arcsin ( \frac{1}{2} ( x - 2 ) ) ) sin ^{3} ( arcsin ( \frac{1}{2} ( x - 2 ) ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (\frac{1}{2} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 2))))) : - \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1))

= - \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1))

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1)) + C

\frac{d x}{d t} = 2 \frac{x}{t} - 0.194 t

\frac{\partial}{\partial x} (ln (x^{6} - y^{6}))

\frac{d}{d M} (\frac{2 π r ^{\frac{3}{2}}}{GM})

\frac{d}{d x} (x^{3} - \frac{4}{x})

t an g e n t y = \frac{2 x}{( x + 1 ) ^{2}}, (0, 0)