用换元法求不定积分的题

x＝－1/2×(1－2x)＋1/2,所以
∫xdx/√(5+x-x^2)
＝－1/2×∫(1-2x)dx/√(5+x-x^2)＋1/2×∫dx/√(5+x-x^2)
＝－1/2×∫1/√(5+x-x^2) d(5+x-x^2) ＋ 1/2×∫1/√[21/4－(x-1/2)^2]
＝－√(5+x-x^2)＋1/2×arcsin[(2x-1)/√(21)]＋C