分类：教育/科学-学习帮助
问题：求下列不定积分:(1)∫[(cos2x)/(sin^2x)]dx 
(1)∫[(c 2x)/(sin^2 x)] dx(2)∫e^x[a^x -(e^-x)/(1-x)] dx(3)∫[(2*3^x)-(5*2^x)]/3^x dx希望能有尽量详细的步骤！谢谢！！
回答：解:(!) ∫[(cos 2x)/(sin^2 x)] dx
      =∫[(cos 2x)(csc^2 x)]dx
而  (csc^2 x)dx =dcotx
故用分步积分得  原式= (cos 2x)*cotx-∫(-2sin2x)cotxdx
                    =(cos 2x)*cotx+4∫sinxcosxcotxdx
=cos2x*cotx+4∫cos^2 xdx=cos2x*cotx+2∫(1+cos2x)dx
=cos2x*cotx+2x+sin2x+c
(2)∫e^x[a^x -(e^-x)/(1-x)] dx
=∫[(ea)^x+1/(x-1)]dx=(ea)^x/ln(ea)+ln(x-1)+c
(3) ∫[(2*3^x)-(5*2^x)]/3^x dx
=∫[2-5*(2/3)^x]dx=2x-5*(2/3)^x/ln(2/3)+c