Verified answer

Notice that denominator has derivative = 3x², which is just 3 times numerator.

So we'll use a substitution, letting u = denominator, du = factor of numerator.

u = x³ - 3

du = 3x² dx

∫ x²/(x³ - 3) dx

= 1/3 ∫ 3x²/(x³ - 3) dx

= 1/3 ∫ 1/(x³ - 3) * 3x² dx

= 1/3 ∫ 1/u du

= 1/3 ln|u| + C

= 1/3 ln|x³ - 3| + C

-- Ματπmφm --

Let u = x³ - 3. Then du = 3x² dx. So x² dx = (1/3) du. The original integral, when substituted, becomes

INT( (1/3) du/u) = (1/3) ln u = (1/3) (x³ - 3) + C.

Derivative over function integrates to the natural logarithm of the function.

Integrate this expression using the above rule:

∫ x² / (x³ - 3) dx = ∫ 3x² / (x³ - 3) dx / 3

∫ x² / (x³ - 3) dx = ln|x³ - 3| / 3 + C

=(1/3) ∫3x^2/(x^3-3)=(1/3)ln(x^3-3)