∫[1/(2 cos x) + 2/(1 + x²) - x/3] dx= ∫ (1/2)(sec x) + 2[1/(1 + x²)] - x/3 dx= (1/2)∫ sec x + 4[1/(1 + x²)] - 2x/3 dx把他们分出来（因为各有各的解决方法）,∫ sec x dx,乘于 (sec x + ta...根号是从哪里到哪里打个比例，
(1/2) ln x
= ln x^(1/2)
= ln √x