11) Find the following indefinite integrals.

∫x/(x+9)^(1/2)dx

Let u^2 = x+9

Then √(x+9) = u
x = u^2-9
dx = 2u du

Then the integral becomes

∫(u^2-9)/u (2u du)
= ∫2(u^2-9) du
= 2/3 u^3 - 18u
= 2u/3 (u^2-27)
= 2/3 √(x+9) (x-18)

For other problems, you can check your results at

http://www.wolframalpha.com/input/?i=%E2%88%ABx%2F%28x%2B9%29^%281%2F2%29dx

Even if you have trouble with the solution, someimes seeing the answer can give you a hint where to start.

To find the indefinite integral ∫x/(x+9)^(1/2)dx, we can use a technique called substitution.

Let's start by making a substitution: u = x+9. Then, we can express dx in terms of du: du = dx.

Now, let's solve for x in terms of u. Subtracting 9 from both sides of the equation, we have x = u - 9.

Substituting these expressions into the integral, we get:
∫(u-9)/u^(1/2) du.

Next, let's simplify the integrand by expanding the numerator:
∫(u/u^(1/2)) - (9/u^(1/2)) du.

Now, we can split this integral into two separate integrals:
∫(u/u^(1/2)) du - ∫(9/u^(1/2)) du.

Simplifying further, we have:
∫u^(1/2) du - 9 * ∫u^(-1/2) du.

Integrating each term individually, we get:
(2/3)u^(3/2) - 9 * 2u^(1/2) + C.

Finally, substituting back u = x+9, we have:
(2/3)(x+9)^(3/2) - 18(x+9)^(1/2) + C, where C is the constant of integration.

Therefore, the indefinite integral of x/(x+9)^(1/2)dx is (2/3)(x+9)^(3/2) - 18(x+9)^(1/2) + C.