<s>[INST] Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other way out.Please help... 
 [/INST] $\displaystyle I = \int \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{5x^4+4x^5}{x^{10}\cdot \left(1+x^{-4}+x^{-5}\right)}dx = \int\frac{(5x^{-6}+4x^{-5})}{(1+x^{-4}+x^{-5})^2}dx$
Now Put $(1+x^{-4}+x^{-5}) = t\;,$ Then $\displaystyle \left(4x^{-5}+5x^{-6}\right)dx = -dt$
So $\displaystyle I = -\int\frac{1}{t^2} = \frac{1}{t}+\mathcal{C} = \frac{x^5}{x^5+x+1}+\mathcal{C}$
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