We will use integration by substitution to solve this integral.

Let 1-x^2=sin^2(u)

This means that:

x^2=1-sin^2(u)=cos^2(u) (using the trigonometric identity sin^2(x)+cos^2(x)=1). Therefore, x=cos(u).

If we sub these results into the original integral we get

∫3cos(u)/√sin^2(u)

Before we can solve this we also need to change the dx at the end of the equation to a du.

We calculate dx/du=-sin(u).

Therefore, we can rewrite the integral as:

∫3cos(u)/sin(u)*(-sin(u))du.

Cancelling out both sin(u)'s, we get:

∫-3cos(u)du = -3sin(u) + C

Now we need to write the solution in terms of x:

Remember that 1-x^2=sin^2(u), therefore sin(u)=√(1-x^2).

So the solution is -3√(1-x^2)+C.

Derive the following: f(x)=(96/x^2)+kx