答：
∫[x³/(1-x²)^(3/2)]dx
=∫[x³/(1-x²)√(1-x²)]dx         设x=sint,-π/2<t<π/2
=∫(sin³t/cos³t)d(sint)
=∫(sin³t/cos²t)dt
=-∫(sin²t/cos²t)d(cost)
=∫[(cos²t-1)/cos²t]d(cost)
=cost+1/cost+C
=√(1-x²)+1/√(1-x²)+C
=(2-x²)/√(1-x²)+Cx³/(1-x²)^(3/2)本来就限定了-10作为分母不能为0