Find the Integral?

Question

Find the antiderivative

⌠ 8x / √(x^2+2)
⌡

Answer 1

Let u = x² + 2
du = 2x dx
4 du = 8x dx
I = 4 ∫ du / u^(1/2)
I = 4 ∫ u^(-1/2) du
I = 8 u^(1/2) + C
I = 8 (x² + 2)^(1/2) + C

Answer 2

∫ 8 x /√(x^2 + 2]) dx

Since the derivative x^2+2 appears in the numerator you can use u substitution:
u = x^2 + 2
du = 2x dx
dx = du/2x

After Substituting you obtain
∫ 8 x /√u) du/2x

Simplifying
∫ 4 /√u) du
4∫ 1/√u) du
4∫ u^(-1/2) du

Integrating and simpifying
4 [ 2u^(1/2)] +C
8u^(1/2) + C

Putting it back in terms of x
8√(x^2+2) + C

Answer 3

put (x^2 +2) =t

then 2xdx = dt

then the integral becomes 4dt/sqrt(t) which is easy