integral of (x^2)/(sqrt(21+4x-x^2))

Lösung

\int \frac{x ^{2}}{21 + 4 x - x ^{2}} d x

\frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21 + C

Schritte zur Lösung

\int \frac{x ^{2}}{21 + 4 x - x ^{2}} d x

Vervollständige das Quadrat

21 + 4 x - x^{2} : - (x - 2)^{2} + 25

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= \int (5 sin (v) + 2)^{2} d v

Multipliziere aus

(5 sin (v) + 2)^{2} : 25 sin^{2} (v) + 20 sin (v) + 4

= \int 25 sin^{2} (v) + 20 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 25 sin^{2} (v) d v + \int 20 sin (v) d v + \int 4 d v

\int 25 sin^{2} (v) d v = \frac{25}{2} (v - \frac{1}{2} sin (2 v))

\int 20 sin (v) d v = - 20 cos (v)

\int 4 d v = 4 v

= \frac{25}{2} (v - \frac{1}{2} sin (2 v)) - 20 cos (v) + 4 v

Ersetze zurück

= \frac{25}{2} (arcsin (\frac{1}{5} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 2)))) - 20 cos (arcsin (\frac{1}{5} (x - 2))) + 4 arcsin (\frac{1}{5} (x - 2))

Vereinfache

\frac{25}{2} (arcsin (\frac{1}{5} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 2)))) - 20 cos (arcsin (\frac{1}{5} (x - 2))) + 4 arcsin (\frac{1}{5} (x - 2)) : \frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21

= \frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21

Füge eine Konstante zur Lösung hinzu

= \frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21 + C

in v erse l a pl a ce \frac{( s + 9 )}{( s ^{2} + 6 s + 34 )}

\frac{d y}{d x} - \frac{2}{x} y = \frac{3}{x ^{2}} y^{4}

\int_{1}^{2} 13 4 x^{2} - 3 d x

d er i v a t i v e y = (4 x - 5) 2

\int 7 cot^{4} (5 t) d t