x^3/sqrt(4+x^2)dx Find the integral above using integration by parts: dv = x/sqrt(4+x^2)dx

Question

For Calculus BC, integration by parts

Answer 1

∫ [x³ / √(x²+ 4)]dx = ∫ [x² (x / √(x² + 4)]dx

Let u = x² and dv = (x / √(x²+4))dx.

Then du = 2xdx and v = ∫dv = ∫ [x / √(x²+4)]dx Let w = x²+4, so dw = 2xdx. xdx = (1/2)dw

Therefore v = (1/2)∫ w^-1/2dw = √w = √(x²+4)

So, ∫ [x³ / √(x² + 4)]dx = uv - ∫vdu = x²√(x²+4) - 2∫x√(x²+4)dx = x²√(x²+4) - ∫w^1/2dw =

x²√(x²+4) - (2/3)w^3/2 + C = x²√(x²+4) - (2/3)(x²+4)^3/2 + C