¿Integral x raiz 2x + 1 dx?

Hola,

∫ x √(2x + 1) dx =

pongamos (el radical):

√(2x + 1) = u

2x + 1 = u²

2x = u² - 1

x = (1/2)(u² - 1)

dx = (1/2) 2u du = u du

obteniendo, por sustitución:

∫ x √(2x + 1) dx = ∫ (1/2)(u² - 1) u u du =

(desarrollando)

∫ (1/2)u² (u² - 1) du =

∫ [(1/2)u⁴ - (1/2)u²] du =

(partiendo en dos integrales y sacando las constantes)

(1/2) ∫ u⁴ du - (1/2) ∫ u² du =

(1/2) [1/(4+1)] u⁴ ⁺ ¹ - (1/2) [1/(2+1)] u² ⁺ ¹ + C =

(1/2)(1/5)u⁵ - (1/2)(1/3)u³ + C =

(sacando (1/2)u³)

(1/2)u³ [(1/5)u² - (1/3)] + C =

(1/2)u³ [(3u² - 5)/15] + C =

(1/30)(3u² - 5)u³ + C

sustituyamos nuevamente √(2x + 1) a u:

(1/30) [3√(2x + 1)² - 5] √(2x + 1)³ + C =

(1/30) [3(2x + 1) - 5] √(2x + 1)³ + C =

(1/30)(6x + 3 - 5)√(2x + 1)³ + C =

(1/30)(6x - 2)√(2x + 1)³ + C =

(1/30) 2(3x - 1)√(2x + 1)³ + C =

concluyendo con:

(1/15)(3x - 1)√(2x + 1)³ + C

espero que sea de ayuda

¡Saludos!