设t=tanx,则dt=d(tanx)
∴原式=∫dx/(2sin²x+3cos²x)
=∫dx/[cos²x(2sin²x/cos²x+3)]
=∫sec²xdx/(2sin²x/cos²x+3)
=∫d(tanx)/(2tan²x+3)
=∫dt/(2t²+3)
=1/3∫dt/(2t²/3+1)
=1/√6∫d(√(2/3)t)/[(√(2/3)t)²+1]
=1/√6arctan[√(2/3)tanx]+C (C是积分常数)