- 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！！
- 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