integral de 1 a 3 de sqrt(1+t^2)

Solución

\int_{1}^{3} 1 + t^{2} d t

\frac{2 ( - 1 + 3 5 ) - ln ( 1 + 2 ) + ln ( 3 + 10 )}{2}

Notación decimal

4.50484 \dots

Pasos de solución

\int_{1}^{3} 1 + t^{2} d t

Aplicar integración por sustitución trigonométrica

: \int_{\frac{π}{4}}^{a r c t a n (3)} sec^{3} (u) d u

= \int_{\frac{π}{4}}^{a r c t a n (3)} sec^{3} (u) d u

Aplicar la reducción de integrales

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

\int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u = ln (3 + 10) - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Simplificar

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Calcular los limites:

\frac{3 5 - 1}{2}

= \frac{3 5 - 1}{2} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Simplificar

= \frac{2 ( - 1 + 3 5 ) - ln ( 1 + 2 ) + ln ( 3 + 10 )}{2}

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