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

Solución

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

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

Notación decimal

1.07357 \dots

Pasos de solución

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

Completar el cuadrado

x^{2} + 2 x : (x + 1)^{2} - 1

= \int_{0}^{1} (x + 1)^{2} - 1 d x

Aplicar integración por sustitución

: \int_{1}^{2} u^{2} - 1 d u

= \int_{1}^{2} u^{2} - 1 d u

Aplicar integración por sustitución trigonométrica

: \int_{0}^{\frac{π}{3}} tan^{2} (v) sec (v) d v

= \int_{0}^{\frac{π}{3}} tan^{2} (v) sec (v) d v

Re-escribir usando identidades trigonométricas

= \int_{0}^{\frac{π}{3}} (- 1 + sec^{2} (v)) sec (v) d v

Expandir

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

= \int_{0}^{\frac{π}{3}} - sec (v) + sec^{3} (v) d v

Aplicar la regla de la suma:

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

= - \int_{0}^{\frac{π}{3}} sec (v) d v + \int_{0}^{\frac{π}{3}} sec^{3} (v) d v

\int_{0}^{\frac{π}{3}} sec (v) d v = ln (2 + 3)

\int_{0}^{\frac{π}{3}} sec^{3} (v) d v = 3 + \frac{1}{2} ln (2 + 3)

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

Simplificar

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

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

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