evaluate the integral 2x/((9+x^2)^1/2) between -2 and 0

One approach would be to use integration by parts, as we have two functions of x within the integral. This approach would get you to the answer but would be more difficult and convoluted than necessary. One could use integration via substitution to evaluate this integral, which will be much simpler and nicer to compute. To begin with we need to introduce a dummy variable, say u. We want to choose u to make the integral easier and so that one of the functions of x can be canceled. So let u = 9 + x^2. We want to write the integral in terms of u now and replace all of the x terms with u. Let's begin by replacing the dx at the end of the integral. We can do this by differentiating u = 9 + x^2. Now du/dx = 2x and dx/du = 1/2x, so dx = du/2x .Don't forget to change the limits now though: u = 9 + (0)^2 = 9u = 9 + (-2)^2 =13 , but 13 is still the lower limit here as -2 was the lower limit, despite the fact that 13 is greater than 9.Now we can rewrite the integral just in terms of u. So we have (2x/(u^1/2))*(du/2x) between 13 and 9. We can see that the 2x's cancel each other out which was our goal so now the integral is just one function in terms of u.To integrate we follow the formula: add 1 to the power; divide by the power, and divide by the differential of the base.Hence, we have 2u^1/2 evaluated at 13 and 9.Inserting the number we get 2(9^1/2) - 2(13^1/2) = 6 - 2(13^1/2).