find the integral of square root of 1+x^3 dx or sqrt(1+ x^3)dx

much too nasty to work out here, try

http://integrals.wolfram.com/index.jsp?expr=sqrt(1%2Bx^3)&random=false

To find the integral of sqrt(1+x^3)dx, we can use a technique called substitution.

Let's make the substitution u = 1 + x^3. Then, we can find du/dx by taking the derivative of both sides with respect to x:

du/dx = 3x^2

To solve for dx, we can rearrange the equation as dx = du/(3x^2).

Now, we can substitute these values into the integral:

∫sqrt(1 + x^3)dx = ∫sqrt(u) * dx/(3x^2)

Next, we need to express everything in terms of u. We already have dx in terms of u, so we just need to find x in terms of u. We can rearrange u = 1 + x^3 to solve for x:

x^3 = u - 1

Taking the cube root of both sides gives us:

x = (u - 1)^(1/3)

Now, we can substitute x = (u - 1)^(1/3) back into the integral:

∫sqrt(1 + x^3)dx = ∫sqrt(u) * (du/(3 * (u - 1)^(2/3))

Simplifying the expression gives:

∫sqrt(1 + x^3)dx = (1/3) * ∫sqrt(u) / (u - 1)^(2/3) * du

Now, we can integrate the expression:

∫sqrt(1 + x^3)dx = (1/3) * ∫u^(-1/2) * (u - 1)^(-2/3) du

This integral can be solved using standard rules of integration or by using techniques such as partial fractions or power rule.

After evaluating the integral, you will have the final result.