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∫(√x - 1/2 x + 2/√x) dx =

∫√x dx - 1/2 ∫x dx + 2∫1/√x dx =

2/3 √x³ - 1/2 x²/2 + 2*2√x + C =

2/3 x√x - 1/4 x² + 4√x + C =

(2/3 x + 4)√x - 1/4 x² + C <===========

∫(1+√x)^2/√x dx =

∫(1+2√x+x)/√x dx =

∫1/√x + 2√x/√x + x/√x) dx =

∫1/√x dx + 2∫dx + ∫√x dx =

2√x + 2x + 2/3 √x³ + C =

2√x + 2x + 2/3 x√x + C =

(2+2/3 x)√x + 2x + C <========

Suerte

1. ∫(√x - 1/2 x + 2/√x) dx =

= ∫(√x)dx - ∫(1/2 x)dx + ∫(2/√x)dx =

= ∫x^(1/2),dx - (1/2).∫x.dx + 2∫x^(-1/2).dx =

= (2/3).x^(3/2) - (1/4).x^2 + 4.x^(1/2) + C

2. ∫ ( 1 + √x)^2/√x dx =

= ∫ ( 1 +2√x+x)/√x dx =

= ∫ (1/√x + 2√x/√x + x/√x) dx =

= ∫ [x^(-1/2) + 2 + x^(1/2)] dx =

= ∫x^(-1/2) dx + 2∫dx + ∫x^(1/2) dx =

= 2.x^(1/2) + 2x + (2/3) x^(3/2) + C

Hola

Suponemos

1)

∫(√x - (1/2) x + (2/√x)) dx = ∫( x^(1/2) - (1/2) x^1 + 2 x^(-1/2)) dx

∫(√x - (1/2) x + (2/√x)) dx = (1/(3/2)) x^(3/2) - (1/2) (1/2) x^2 +

+ 2 (1/(1/2)) x^(1/2) + C

∫(√x - (1/2) x + (2/√x)) dx = (2/3) x^(3/2) - (1/4) x^2 + 4 x^(1/2) + C

2)

∫ (( 1 + √x)^2/√x) dx

sustituimos

u = √x

x = u^2

dx = 2 u du

1 + √x = 1 + u

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

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

∫ (( 1 + √x)^2/√x) dx = (2/3) (1 + u)^3 + C

∫ (( 1 + √x)^2/√x) dx = (2/3) (1 + √x )^3 + C