integral de 3sqrt(x^2-1)

Solução

\int 3 x^{2} - 1 d x

3 (\frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1) + C

Passos da solução

\int 3 x^{2} - 1 d x

Remover a constante:

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

= 3 \cdot \int x^{2} - 1 d x

Aplicar integração por substituição trigonométrica

: \int tan^{2} (u) sec (u) d u

= 3 \cdot \int tan^{2} (u) sec (u) d u

Reeecreva usando identidades trigonométricas

= 3 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expandir

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 3 \cdot \int - sec (u) + sec^{3} (u) d u

Aplicar a regra da soma:

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

= 3 (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 3 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substituir na equação

u = arcsec (x)

= 3 (- ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ + \frac{1}{2} sec (arcsec (x)) tan (arcsec (x)) + \frac{1}{2} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣)

Simplificar

3 (- ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ + \frac{1}{2} sec (arcsec (x)) tan (arcsec (x)) + \frac{1}{2} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣) : 3 (\frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1)

= 3 (\frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1)

Adicionar uma constante à solução

= 3 (\frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1) + C

\frac{d}{d x} (e^{x + 3})

\int_{0}^{1} \frac{- 10 y}{e ^{2 y}} d y

x \to 3 + lim (\frac{( x - 8 )}{x - 3})

x \to 0 + lim ((x) (e^{\frac{1}{x}}))

y = \frac{x ^{2}}{4 + x}