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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

3 - 2 x - x^{2} : - (x + 1)^{2} + 4

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = x + 1

ein

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

Multipliziere aus

- (x + 1)^{2} + 4 : - x^{2} - 2 x + 3

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

Füge eine Konstante zur Lösung hinzu

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

\int_{2}^{\frac{1}{x}} arctan (t) d t

\int_{- 5}^{0} (4 + 25 - x^{2}) d x

x \to 8 lim (\frac{x + 9}{x - 8})

y^{^{''}} + 256 y = 7 cos (5 t)

in v erse l a pl a ce \frac{10}{s ^{2} + 25} + \frac{4}{s - 3}