integral de 0 a 5 de sqrt(x^2+3x+1)

Solução

\int_{0}^{5} x^{2} + 3 x + 1 d x

\frac{- 5 ln ( \frac{333 + 52 41}{25} ) + 4 ( 13 41 - 3 )}{16}

Decimal

19.03441 \dots

Passos da solução

\int_{0}^{5} x^{2} + 3 x + 1 d x

Completar o cuadrado

x^{2} + 3 x + 1 : (x + \frac{3}{2})^{2} - \frac{5}{4}

= \int_{0}^{5} (x + \frac{3}{2})^{2} - \frac{5}{4} d x

Aplicar integração por substituição

= \int_{\frac{3}{2}}^{\frac{13}{2}} \frac{4 u ^{2} - 5}{2} d u

Remover a constante:

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

= \frac{1}{2} \cdot \int_{\frac{3}{2}}^{\frac{13}{2}} 4 u^{2} - 5 d u

Aplicar integração por substituição trigonométrica

= \frac{1}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} \frac{5}{2} tan^{2} (v) sec (v) d v

Remover a constante:

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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} tan^{2} (v) sec (v) d v

Reeecreva usando identidades trigonométricas

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} (- 1 + sec^{2} (v)) sec (v) d v

Expandir

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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} - sec (v) + sec^{3} (v) 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

= \frac{1}{2} \cdot \frac{5}{2} (- \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec (v) d v + \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec^{3} (v) d v)

\int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec (v) d v = ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)

\int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec^{3} (v) d v = \frac{13 41 - 3}{5} + \frac{1}{2} (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5))

= \frac{1}{2} \cdot \frac{5}{2} (- (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)) + \frac{13 41 - 3}{5} + \frac{1}{2} (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)))

Simplificar

\frac{1}{2} \cdot \frac{5}{2} (- (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)) + \frac{13 41 - 3}{5} + \frac{1}{2} (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5))) : \frac{- 5 ln ( \frac{333 + 52 41}{25} ) + 4 ( 13 41 - 3 )}{16}

= \frac{- 5 ln ( \frac{333 + 52 41}{25} ) + 4 ( 13 41 - 3 )}{16}