求不定积分dx/根号(1+x-x^2)请老师详细一点谢谢

原式=∫dx/√-(x^2-x+1/4-1/4-1)
=∫dx/√[5/4-(x-1/2)^2]
=∫dx/√[(√5/2）^2-(x-1/2)^2]
=arcsin[(x-1/2)/(√5/2)]+C
=arcsin[(2x-1)/√5]+C.
利用三角代换证明公式∫dx/√（a^2-x^2)=arcsin(x/a)+C.