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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

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

= 4 \cdot \int \frac{x ^{2}}{4 x - x ^{2}} d x

Vervollständige das Quadrat

4 x - x^{2} : - (x - 2)^{2} + 4

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= 4 \cdot \int (2 sin (v) + 2)^{2} d v

Multipliziere aus

(2 sin (v) + 2)^{2} : 4 sin^{2} (v) + 8 sin (v) + 4

= 4 \cdot \int 4 sin^{2} (v) + 8 sin (v) + 4 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

= 4 (\int 4 sin^{2} (v) d v + \int 8 sin (v) d v + \int 4 d v)

\int 4 sin^{2} (v) d v = 2 v - sin (2 v)

\int 8 sin (v) d v = - 8 cos (v)

\int 4 d v = 4 v

= 4 (2 v - sin (2 v) - 8 cos (v) + 4 v)

Ersetze zurück

= 4 (2 arcsin (\frac{1}{2} (x - 2)) - sin (2 arcsin (\frac{1}{2} (x - 2))) - 8 cos (arcsin (\frac{1}{2} (x - 2))) + 4 arcsin (\frac{1}{2} (x - 2)))

Vereinfache

4 (2 arcsin (\frac{1}{2} (x - 2)) - sin (2 arcsin (\frac{1}{2} (x - 2))) - 8 cos (arcsin (\frac{1}{2} (x - 2))) + 4 arcsin (\frac{1}{2} (x - 2))) : 4 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x)

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (8 x e^{2 x})

\int e^{x} tan^{4} (e^{x}) d x

n = 1 \sum \infty \frac{n}{n + 1}

\int - t^{3} d t

\frac{\partial}{\partial y} (- z)