integral of (2x+5)/(sqrt(16-6x-x^2))

Lösung

\int \frac{2 x + 5}{16 - 6 x - x ^{2}} d x

- 2 - x^{2} - 6 x + 16 - arcsin (\frac{1}{5} (x + 3)) + C

Schritte zur Lösung

\int \frac{2 x + 5}{16 - 6 x - x ^{2}} d x

Vervollständige das Quadrat

16 - 6 x - x^{2} : - (x + 3)^{2} + 25

= \int \frac{2 x + 5}{- ( x + 3 ) ^{2} + 25} d x

Wende U-Substitution an

= \int \frac{2 u - 1}{- u ^{2} + 25} d u

Multipliziere aus

\frac{2 u - 1}{- u ^{2} + 25} : \frac{2 u}{- u ^{2} + 25} - \frac{1}{- u ^{2} + 25}

= \int \frac{2 u}{- u ^{2} + 25} - \frac{1}{- u ^{2} + 25} 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} + 25} d u - \int \frac{1}{- u ^{2} + 25} d u

\int \frac{2 u}{- u ^{2} + 25} d u = - 2 - u^{2} + 25

\int \frac{1}{- u ^{2} + 25} d u = arcsin (\frac{1}{5} u)

= - 2 - u^{2} + 25 - arcsin (\frac{1}{5} u)

Setze in

u = x + 3

ein

= - 2 - (x + 3)^{2} + 25 - arcsin (\frac{1}{5} (x + 3))

Multipliziere aus

- (x + 3)^{2} + 25 : - x^{2} - 6 x + 16

= - 2 - x^{2} - 6 x + 16 - arcsin (\frac{1}{5} (x + 3))

Füge eine Konstante zur Lösung hinzu

= - 2 - x^{2} - 6 x + 16 - arcsin (\frac{1}{5} (x + 3)) + C

\int x^{3} x^{2} + 31 d x

x \to 0 lim (I n (x))

\int_{- 2}^{3} (x - 6) - (x^{2} - 12) d x

x \to 1 lim (\frac{8}{x ^{3} - 1})

x \to 0 lim (7 x)