Evaluate the integral ∫2x√(x^2 +1) dx

To evaluate the integral ∫2x√(x^2 +1) dx, we can use a u-substitution.

Let u be equal to x^2 + 1. Then, du/dx = 2x, which means dx = du/(2x).

Substituting these values into the integral, we have:

∫2x√(x^2 +1) dx = ∫2x√u * (du/(2x))

The x in the numerator and denominator cancels out:

= ∫√u du

Now, we can integrate with respect to u:

= (2/3) u^(3/2) + C

Substituting back u = x^2 + 1, we get:

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

Therefore, the integral ∫2x√(x^2 +1) dx equals (2/3) (x^2 + 1)^(3/2) + C.

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