Calculus help..?

Question

Integration: Find (2x^3 = 3x)square root of (x^4+3x^2)dx

Thanks

Answer 1

First, I'm going to assume that your equals sign
is a typo for a + sign.
Note that the integral is 1/2 ∫ (4x^3+6x)√(x^4 + 3x^2) dx

Now let u = x^4 + 3x^2, du = (4x^3 + 6x)dx.

So we get (1/2)*∫ √u du (3/4)u^(3/2) + C

= (3/4)*(x^4 + 3x^2)^(3/2) + C.

Answer 2

I don't understand your question. What is the 2x^3 = 3 part?