Q-1   (5x⁴+4x⁵)/(x⁵+x+1)²

^Q-2cos²x/1+tanx

cosx^2/(1+tanx)=(1-tanx^2)/(1+tanx^2)(1+tanx)=(1-tanx)/(1+tan^2x)=(1-tanx)/secx^2=cosx^2-tanx/secx=(cos2x+1)/2-1/2(2sinxcosx)=1/4sin2x+x-1/2sin2x=1/4sin2x+x-1/4(-cos2x)=1/4sin2x+x+1/4sin2x

(5x^4+4x^5)/(x^5+x+1)^2=x^5(5/x+4)/x^10(1+1/x^4+1/x^5)=(5/x^6+4/x^5)/(1+1/x^4+1/x^5)                 let (1/x^4+1+1/x^5)=t    hence (5/x^6+4/x^5)dx=dt     hence,  1/t^2  dt =-1/t =-1/(1+1/x^4+1/x^5)  is the ANSWER.             PLEASE RATE IF SATISFIED.

Good method angel.I did a lot of problems like this but my brain didn't work in this case.Anyway thanx,I have rated u.

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