Finding an indefinite integral

Question

∫ (x+5)/√(9-(x-3)^2) dx

I know I need to split the numerator into (x-3) +8, but after that I get stuck.

Answer 1

Using your first step as a starting point, we have

∫(x-3)/√(9-(x-3)²)dx + 8∫1/√(9-(x-3)²)dx

For the first integral ∫(x-3)/√(9-(x-3)²)dx, let u = 9-(x-3)², du = -2(x-3)dx, or -1/2*du = (x-3)dx, yielding∫

-1/2 * ∫u^-1/2du = -1/2 * 2 u^1/2 = -u^1/2 = -(9-(x-3)²)^1/2. I'll add the constant of integration after the second integral.

For the second integral 8∫1/√(9-(x-3)²)dx perform the substitution u = (x-3)/3, du = dx/3 yielding

8∫du/√[9-(3u)²] = 8/3 * ∫du/3√1-u² = 8/9 * arcsin(u) + C = 8/9 * arcsin((x-3)/3) + C

Final answer: -(9-(x-3)²)^1/2 + 8/9 * arcsin((x-3)/3) + C

Answer 2

I'd be happy to help you to the next step:

∫ x−3/√9−(x−3)² dx + 8 ∫1/√9−(x−3)² dx

Substitute u=9−(x−3)² ⟶ du/dx=−2(x−3) ⟶ dx=−1/2(x−3)du = -1/2 ∫ 1/√u du

next, apply the power rule.