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integral de 0 a 2 de sqrt(5x^2+4x+29)

Solución

\int_{0}^{2} 5 x^{2} + 4 x + 29 d x

Solución

\frac{1}{10 5} (282 (\frac{2 285}{47} - \frac{145}{141}) + 141 ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141}))

+1

Notación decimal

12.53040 \dots

Pasos de solución

\int_{0}^{2} 5 x^{2} + 4 x + 29 d x

Completar el cuadrado

5 x^{2} + 4 x + 29 : 5 (x + \frac{2}{5})^{2} + \frac{141}{5}

= \int_{0}^{2} 5 (x + \frac{2}{5})^{2} + \frac{141}{5} d x

Aplicar integración por sustitución

: \int_{\frac{2}{5}}^{\frac{12}{5}} \frac{25 u ^{2} + 141}{5} d u

= \int_{\frac{2}{5}}^{\frac{12}{5}} \frac{25 u ^{2} + 141}{5} d u

\frac{25 u ^{2} + 141}{5} = \frac{1}{5} 25 u^{2} + 141

= \int_{\frac{2}{5}}^{\frac{12}{5}} \frac{1}{5} 25 u^{2} + 141 d u

Sacar la constante:

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

= \frac{1}{5} \cdot \int_{\frac{2}{5}}^{\frac{12}{5}} 25 u^{2} + 141 d u

Aplicar integración por sustitución trigonométrica

: \int_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} \frac{141}{5} sec^{3} (v) d v

= \frac{1}{5} \cdot \int_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} \frac{141}{5} sec^{3} (v) d v

Sacar la constante:

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

= \frac{1}{5} \frac{141}{5} \cdot \int_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} sec^{3} (v) d v

Aplicar la reducción de integrales

= \frac{1}{5} \frac{141}{5} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} sec (v) d v

\int_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} sec (v) d v = ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141})

= \frac{1}{5} \frac{141}{5} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} + \frac{1}{2} ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141})

Simplificar

\frac{1}{5} \frac{141}{5} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} + \frac{1}{2} ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141}) : \frac{1}{10 5} 282 [\frac{sec ( v ) tan ( v )}{2}]_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} + 141 ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141})

= \frac{1}{10 5} 282 [\frac{sec ( v ) tan ( v )}{2}]_{a r c t a n (\frac{2}{141})}^{a r c t a n (\frac{4 3}{47})} + 141 ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141})

Calcular los limites:

\frac{2 285}{47} - \frac{145}{141}

= \frac{1}{10 5} (282 (\frac{2 285}{47} - \frac{145}{141}) + 141 ln (\frac{12 145 + 5 1653 - 24 - 2 285}{141}))

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