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

Solución

\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

Pasos de solución

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

Completar el cuadrado

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

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

Aplicar integración por sustitución

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

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

Aplicar integración por sustitución trigonométrica

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

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

Expandir

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

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

Aplicar la regla de la suma:

\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

Sustitución hacia atrás

= \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))

Simplificar

\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

Agregar una constante a la solución

= \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

\frac{d}{d x} (\frac{\frac{d}{d x} ( e ^{x} )}{1 + x})

\int 2 x (8 x^{2} - 20)^{2} d x

x \to 1 + lim (1 + \frac{1}{x - 1})

\int 7 x + 5 d x

f (x) = sin (x^{5})