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integral de x^2sqrt(3+2x-x^2)

Solução

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

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

Passos da solução

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

Completar o cuadrado

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

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

Aplicar integração por substituição

: \int (u + 1)^{2} - u^{2} + 4 d u

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

Aplicar integração por substituição trigonométrica

: \int 4 cos^{2} (v) (2 sin (v) + 1)^{2} d v

= \int 4 cos^{2} (v) (2 sin (v) + 1)^{2} d v

Remover a constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int cos^{2} (v) (2 sin (v) + 1)^{2} d v

Reeecreva usando identidades trigonométricas

= 4 \cdot \int (1 - sin^{2} (v)) (2 sin (v) + 1)^{2} d v

Expandir

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

= 4 \cdot \int 3 sin^{2} (v) + 4 sin (v) - 4 sin^{4} (v) - 4 sin^{3} (v) + 1 d v

Aplicar a regra da soma:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int 3 sin^{2} (v) d v + \int 4 sin (v) d v - \int 4 sin^{4} (v) d v - \int 4 sin^{3} (v) d v + \int 1 d v)

\int 3 sin^{2} (v) d v = \frac{3}{2} (v - \frac{1}{2} sin (2 v))

\int 4 sin (v) d v = - 4 cos (v)

\int 4 sin^{4} (v) d v = - sin^{3} (v) cos (v) + \frac{3}{2} v - \frac{3}{4} sin (2 v)

\int 4 sin^{3} (v) d v = 4 (- cos (v) + \frac{cos ^{3} ( v )}{3})

\int 1 d v = v

= 4 (\frac{3}{2} (v - \frac{1}{2} sin (2 v)) - 4 cos (v) - (- sin^{3} (v) cos (v) + \frac{3}{2} v - \frac{3}{4} sin (2 v)) - 4 (- cos (v) + \frac{cos ^{3} ( v )}{3}) + v)

Substituir na equação

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

Simplificar

4 (\frac{3}{2} (arcsin (\frac{1}{2} (x - 1)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 1)))) - 4 cos (arcsin (\frac{1}{2} (x - 1))) - (- sin^{3} (arcsin (\frac{1}{2} (x - 1))) cos (arcsin (\frac{1}{2} (x - 1))) + \frac{3}{2} arcsin (\frac{1}{2} (x - 1)) - \frac{3}{4} sin (2 arcsin (\frac{1}{2} (x - 1)))) - 4 (- cos (arcsin (\frac{1}{2} (x - 1))) + \frac{cos ^{3} ( arcsin ( \frac{1}{2} ( x - 1 ) ) )}{3}) + arcsin (\frac{1}{2} (x - 1))) : 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}})

= 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}})

Adicionar uma constante à solução

= 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}}) + C

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

\int_{0}^{\frac{π}{2}} x sin (8 x) d x

x \to - 0 lim (\frac{( x + 2 ) ^{2} - 4}{x})

f (x) = \frac{2}{( 3 x ^{2} + 7 ) ^{2}}

f (x) = 6 x - x^{2}