Question

求定积分的题目

求定积分的题目请各位指教第六题

Answer 1

不定积分的原函数求解

Answer 2

x:0->a
∫xdx/((2-x²)√(1-x²))
=1/2∫dx²/((2-x²)√(1-x²))
=-1/2∫d(1-x²)/((2-x²)√(1-x²)) 设1- x²=t
=-∫d√(1-x²)/(1+1-x²)
=-∫d√(1-x²)/(1+1-x²)
=-arctan√(1-x²)
=arctan1-arctan√(1-a²)
=π/4-arctan√(1-a²)