integral de ((x^2-16)^{3/2})/(x^3)

Solução

\int \frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{3}} d x

\frac{16 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{4}} - 6 arcsec (\frac{1}{4} x) + 3 sin (2 arcsec (\frac{1}{4} x)) + \frac{( x ^{2} - 16 ) ^{\frac{5}{2}}}{x ^{4}} + C

Passos da solução

\int \frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{3}} d x

Aplicar integração por substituição trigonométrica

= \int 4 tan^{4} (u) cos^{2} (u) d u

Remover a constante:

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

= 4 \cdot \int tan^{4} (u) cos^{2} (u) d u

Reeecreva usando identidades trigonométricas

= 4 \cdot \int \frac{sin ^{4} ( u )}{cos ^{2} ( u )} d u

Aplicar integração por partes

= 4 (- \int 4 sin^{4} (u) d u + sin^{4} (u) tan (u))

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

= 4 (- (- sin^{3} (u) cos (u) + \frac{3}{2} u - \frac{3}{4} sin (2 u)) + sin^{4} (u) tan (u))

Substituir na equação

u = arcsec (\frac{1}{4} x)

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

Simplificar

4 (- (- sin^{3} (arcsec (\frac{1}{4} x)) cos (arcsec (\frac{1}{4} x)) + \frac{3}{2} arcsec (\frac{1}{4} x) - \frac{3}{4} sin (2 arcsec (\frac{1}{4} x))) + sin^{4} (arcsec (\frac{1}{4} x)) tan (arcsec (\frac{1}{4} x))) : \frac{16 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{4}} - 6 arcsec (\frac{1}{4} x) + 3 sin (2 arcsec (\frac{1}{4} x)) + \frac{( x ^{2} - 16 ) ^{\frac{5}{2}}}{x ^{4}}

= \frac{16 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{4}} - 6 arcsec (\frac{1}{4} x) + 3 sin (2 arcsec (\frac{1}{4} x)) + \frac{( x ^{2} - 16 ) ^{\frac{5}{2}}}{x ^{4}}

Adicionar uma constante à solução

= \frac{16 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{4}} - 6 arcsec (\frac{1}{4} x) + 3 sin (2 arcsec (\frac{1}{4} x)) + \frac{( x ^{2} - 16 ) ^{\frac{5}{2}}}{x ^{4}} + C

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