Question

不定积分习题 要详细过程 拜托了 快啊！！！

Answer 1

∫(sinx+cosx)dx/(sinx-cosx)^3
=∫d(sinx-cosx)/(sinx-cosx)^3
=(-1/2)(1/(sinx-cosx)^2 +C

∫sinxdx/[(cosx)^3(1+(secx)^2)^(1/3)]
=-∫dcosx/[cosx^3(1+secx^2)^(1/3)]
=(1/2)∫dsecx^2/(1+secx^2)^(1/3)
=(1/2)(3/2)(secx^2+1)^(2/3) +C

∫dx/[x^2√(x^2+3)]
x=√3tanu dx=√3dtanu=√3secu^2du tanu=x/√3 cotu=√3/x
1/sinu^2=cotu^2+1=[(3+x^2)/x^2 1/sinu=√(3+x^2)/x
=∫√3secu^2du/[tanu^2√3secu}
=∫cosudu/sinu^2
=-∫dsinu/sinu^2
=1/sinu+C
=√(3+x^2)/x+C
=

Answer 2

S(sinx+cosx)/(sinx-cosx)^3dx
=S1/(sinx+cosx)^3d(-cosx+sinx)
=-1/3(sinx-cosx)^2
写的太麻烦了