Evaluate the following Integral

Question

∫ (2√3x^2 -1 - 2/x)(3x/√3x^2 - 1 + 1/x^2) dx

Answer 1

I think that you need parentheses around 3x² - 1.

∫ [ 2√(3x²-1) - 2/x] [ 3x/√(3x²-1) + 1/x²]dx

Let u = 2√(3x²-1) - 2/x. Then du = (2(1/2)(3x²-1)^-1/2(6x) + 2/x²)dx = 2[3x/√(3x²-1) + 1/x²]dx

So, the given integral is equivalent to (1/2) ∫udu = (1/4)u²+ C = (√(3x² - 1) - 2/x)² + C