(sinx^2+1)/cox^4求不定积分

∫[(sinx)^2+1] dx / [(cosx)^4]
=∫[2-(cosx)^2] dx / [(cosx)^4]
=2∫dx/(cosx)^4 - ∫dx/(cosx)^2
=2∫(secx)^4dx-∫(secx)^2dx
=2∫(secx)^2d(tanx)-tanx
=2[∫(tanx)^2d(tanx) + ∫d(tanx)]-tanx
=2/3 (tanx)^3+tanx+c
=2/3sinx[1-(cosx)^2]/[(cosx)^3]+sinx/cosx+c
=(2/3)sinx / [(cosx)^2 + (1/3)(sinx/cosx)+c