integral of (x+6)/(sqrt(5-4x-x^2))

Lösung

\int \frac{x + 6}{5 - 4 x - x ^{2}} d x

- - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2)) + C

Schritte zur Lösung

\int \frac{x + 6}{5 - 4 x - x ^{2}} d x

Vervollständige das Quadrat

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

= \int \frac{x + 6}{- ( x + 2 ) ^{2} + 9} d x

Wende U-Substitution an

= \int \frac{u + 4}{- u ^{2} + 9} d u

Trigonometrische Substitution anwenden

= \int 3 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

= \int 3 sin (v) d v + \int 4 d v

\int 3 sin (v) d v = - 3 cos (v)

\int 4 d v = 4 v

= - 3 cos (v) + 4 v

Ersetze zurück

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

Vereinfache

- 3 cos (arcsin (\frac{1}{3} (x + 2))) + 4 arcsin (\frac{1}{3} (x + 2)) : - - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2))

= - - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2))

Füge eine Konstante zur Lösung hinzu

= - - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2)) + C

\frac{d}{d t} (ln (cos (\frac{π}{2} - \frac{t}{2})))

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

n = 1 \sum \infty \frac{3 ^{n}}{n}

l a pl a ce t r an s f or m (t - 1) cos (4 (t - 1))

(x - 1) y^{^{'}} + x y = 6 e^{x}