不定积分x^2arcsinx+1/(1-x^2)^2 怎么解

f(x) = x^2arcsinx/√(1-x^2)
f(-x) =-f(x)
∫(-1/2->1/2) ( x^2arcsinx +1 )/√(1-x^2) dx
=∫(-1/2->1/2) dx/√(1-x^2)
=[arcsinx]|(-1/2->1/2)
= π/3追问

第四五行什么意思？没看明白。。。

追答

∫(-1/2->1/2) ( x^2arcsinx +1 )/√(1-x^2) dx
=∫(-1/2->1/2) x^2arcsinx /√(1-x^2) dx +∫(-1/2->1/2) dx/√(1-x^2)
=0+∫(-1/2->1/2) dx/√(1-x^2)
=∫(-1/2->1/2) dx/√(1-x^2)

let
x=siny
dx =cosy dy
x=-1/2 , y=-π/6
x=1/2 , y=π/6

∫(-1/2->1/2) dx/√(1-x^2)
=∫(-π/6->π/6)dy
=π/3

追问

懂了，谢谢~

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