Calculus Trig Sub INTEGRALS HELP!!! Please!

integral x^2/( square root of (4 + x^2), It would help if you could show the work THANKS!

Hello,

∫ [x² /√(4 + x²)] dx =

rewrite it as:

∫ [x² /√(2² + x²)] dx =

let:

x = 2tanθ

tanθ = x/2

dx = 2sec²θ dθ

then, substituting:

∫ [x² /√(2² + x²)] dx = ∫ {(2tanθ)² /√[2² + (2tanθ)²]} 2sec²θ dθ =

∫ [(4tan²θ) /√(2² + 2²tan²θ)] 2sec²θ dθ =

∫ {(4tan²θ) /√[2²(1 + tan²θ)]} 2sec²θ dθ =

(applying the identity 1 + tan²θ = sec²θ)

∫ [(4tan²θ) /√(2²sec²θ)] 2sec²θ dθ =

∫ [(4tan²θ) /(2secθ)] 2sec²θ dθ =

(simplifying and pulling the constant out)

4 ∫ tan²θ secθ dθ =

replace tan²θ with sec²θ - 1:

4 ∫ (sec²θ - 1) secθ dθ =

(expanding)

4 ∫ (secθ sec²θ - secθ) dθ =

(splitting into two integrals)

4 ∫ secθ sec²θ dθ - 4 ∫ secθ dθ

summing up, we have:

4 ∫ tan²θ secθ dθ = 4 ∫ secθ sec²θ dθ - 4 ∫ secθ dθ

let's now integrate ∫ secθ sec²θ dθ by parts, letting:

secθ = u → tanθ secθ dθ = du

sec²θ dθ = dv → tanθ = v

obtaining:

∫ u dv = v u - ∫ v du

4 ∫ tan²θ secθ dθ = 4 [tanθ secθ - ∫ tanθ (tanθ secθ) dθ] - 4 ∫ secθ dθ

4 ∫ tan²θ secθ dθ = 4tanθ secθ - 4 ∫ tan²θ secθ dθ - 4 ∫ secθ dθ

we have the same integral on both sides, thus let's collect it at the left side:

4 ∫ tan²θ secθ dθ + 4 ∫ tan²θ secθ dθ = 4tanθ secθ - 4 ∫ secθ dθ

2 (4 ∫ tan²θ secθ dθ) = 4tanθ secθ - 4 ∫ secθ dθ

hence:

4 ∫ tan²θ secθ dθ = (1/2) (4tanθ secθ - 4 ∫ secθ dθ) =

2tanθ secθ - 2 ∫ secθ dθ =

let's multiply the remaining integrand by (secθ + tanθ) /(tanθ + secθ) (= 1):

2tanθ secθ - 2 ∫ (secθ + tanθ) secθ dθ /(tanθ + secθ) =

(expanding)

2tanθ secθ - 2 ∫ (sec²θ + tanθ secθ) dθ /(tanθ + secθ) =

note that the numerator is the derivative of the denominator:

2tanθ secθ - 2 ∫ d(tanθ + secθ) /(tanθ + secθ) =