∫（2x-1）除以根号x dx ∫cosx dx +∫-2（sinx）^2 乘以cosx dx+∫（sinx)^4乘以c

∫（2x-1）除以根号x dx ∫cosx dx +∫-2（sinx）^2 乘以cosx dx+∫（sinx)^4乘以cosx dx

怎么全是不定积分啊?古怪
∫(2x-1) /√x dx ∫cosx dx +∫-2(sinx)^2 cosx dx+∫(sinx)^4 cosx dx
= [ ∫2√x dx-∫1/√x dx] ∫cosx dx -∫2(sinx)^2 d(sinx)+∫(sinx)^4 d(sinx)
=[(4/3) x^(3/2)+C1-2√x+C2](sinx+C3) - 2(sinx)^3 /3 + C3 + (sinx)^5 / 5 +C4
=[(4/3) x^(3/2) - 2√x + C](sinx + C') - 2(sinx)^3 /3 + (sinx)^5 / 5 +C
其中C,C',C 可以是任意常数