Integration Using U-Substitution

Integrate using u-substitution.

2/(sqrt{3x^2}) dx

∫ 2/√(3x^2) dx
= 2/√3 ∫ 1/x dx
= 2/√3 * ln(x) + C

Edit: using substitution:
∫ 2/√(3x^2) dx
= ∫ 2/[√(3) √(x^2)] dx
= 2/√3 ∫ 1/√(x^2) dx
= 2/√3 ∫ √(x^2)/x^2 dx => sub. u = 1/x^2 , du = -2/x^3 dx , dx = -(x^3 du)/2
= -1/√3 ∫ 1/u du
= -1/√3 * ln(u) + C => sub. back for u = 1/x^2
= -1/√3 * ln(1/x^2) + C
= -1/√3 [ln(1) - ln(x^2)] + C
= -1/√3 [0 - 2ln(x)] + C
= 2/√3 * ln(x) + C

Regards.

You don't need a U-Substitution to integrate: SQRT(3X^2)=SQRT(3)*X
so your integral becomes 2/SQRT(3)*X =[2/SQRT(3)]ln|X| + C