2tanθ secθ - 2 ln |tanθ + secθ| + C
let's now convert this back to x variable, recalling that tanθ = x/2;
hence:
secθ = √(1 + tan²θ) = √[1 + (x/2)²] = √[1 + (x²/4)] = √[(4 + x²) /4] = [√(4 + x²)] /2
then, substituting back:
2tanθ secθ - 2 ln |tanθ + secθ| + C = 2(x/2) {[√(4 + x²)] /2} - 2 ln |(x/2) +
{[√(4 + x²)] /2}| + C =
(x/2)√(4 + x²) - 2 ln |[x + √(4 + x²)] /2| + C =
(applying logarithm properties)
(x/2)√(4 + x²) - 2 [ln |x + √(4 + x²)| - ln (2)] + C =
(x/2)√(4 + x²) - 2 ln |x + √(4 + x²)| + 2 ln (2) + C
being 2 ln (2) a mere constant, we can include it in C, ending with:
∫ [x² /√(4 + x²)] dx = (x/2)√(4 + x²) - 2 ln [x + √(4 + x²)] + C
(the absolute value is no needed any longer since the argument x + √(4 + x²) is always positive)
I hope it's helpful