Question

Evaluate the given indefinite integrals.
∫ (x+2)(2x+3)^1/2

Answer 1

Use Integration by Parts

∫u dv = uv - ∫v du

In your expression,
let dv = (2x+3)^(1/2) dx
v = (1/2)(2/3) (2x+3)^(3/2
v = (1/3) (2x+3)^(3/2)

let u = x+2
du/dx = 1
du = dx

then
∫ (x+2)(2x+3)^1/2
= uv - ∫ v du
= (x+2)(1/3)(2x+3)^(3/2 - ∫ (1/3) (2x+3)^(3/2) dx
= (1/3)(x+2)(2x+3)^(3/2) - (1/15)(2x+3)^(5/2)

= (1/15)(2x+3)^(3/2) [ 5(x+2) - (2x+3) ]
= (1/15)(2x+3)^(3/2) (3x + 7)
= (1/15) (3x+7) (2x+3)^(3/2)

(I checked my answer by differentiating it)