integral of (2x-3)/(sqrt(4x-x^2))

Lösung

\int \frac{2 x - 3}{4 x - x ^{2}} d x

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

Schritte zur Lösung

\int \frac{2 x - 3}{4 x - x ^{2}} d x

Vervollständige das Quadrat

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

= \int \frac{2 x - 3}{- ( x - 2 ) ^{2} + 4} d x

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

= - 2 - u^{2} + 4 + arcsin (\frac{1}{2} u)

Setze in

u = x - 2

ein

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

Vereinfache

- 2 - (x - 2)^{2} + 4 + arcsin (\frac{1}{2} (x - 2)) : - 2 - x^{2} + 4 x + arcsin (\frac{1}{2} x - 1)

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

Füge eine Konstante zur Lösung hinzu

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