Question

Me ayudan con éstos problemas de cálculo integral

Answer 1

∫2x+3/√x²+3x = ∫1/√u du

∫1/√u =u^-1/2

(x^2+3x)=u

[(x^2+3)^-1/2+1]/-1/2+1

[(x^2+3)^1/2]/1/2

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

∫5x^2dx/√4-x^3

5{-1/3[(4-x^3)^-1/2+1]/-1/2+1

5{-1/3[(4-x^3)^1/2]/1/2

5{-1/3[2(4-x^3)^1/2+C

-10/3(4-x^3)^1/2+C

∫x³dx/3√x^4-4

1/6(√x^3-4)+C