Question

∫1/(a²-x²)^5/2dx

Answer 1

∫ dx/√(a² - x²)^5 = ∫ dx/(a² - x²)^(5/2)
(x = a•sinz,dx = a•cosz dz)
= ∫ a•cosz/(a² - a²•sin²z)^(5/2) dz
= ∫ a•cosz/(a²•cos²z)^(5/2) dz
= ∫ a•cosz(a^5 • cos^5z) dz
= (1/a⁴)∫ sec⁴z dz
= (1/a⁴)∫ sec²z d(tanz)
= (1/a⁴)∫ (1 + tan²z) d(tanz)
= (1/a⁴)[tanz + (1/3)tan³z] + C
= (1/a⁴)[x/√(a² - x²) + (1/3) • x³/(a² - x²)^(3/2)] + C
= x(3a² - 2x²)/[3a⁴(a² - x²)^(3/2)] + C