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... 
 A: $\displaystyle I = \int \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{5x^4+4x^5}{x^{10}\cdot \left(1+x^{-4}+x^{-5}\right)}dx = \int\frac{(5x^{-6}+4x^{-5})}{(1+x^{-4}+x^{-5})^2}dx$
Now Put $(1+x^{-4}+x^{-5}) = t\;,$ Then $\displaystyle \left(4x^{-5}+5x^{-6}\right)dx = -dt$
So $\displaystyle I = -\int\frac{1}{t^2} = \frac{1}{t}+\mathcal{C} = \frac{x^5}{x^5+x+1}+\mathcal{C}$
A: Let $\displaystyle \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \frac{d}{dx}\left[\frac{ax+b}{(x^5+x+1)}\right] = \frac{(x^5+x+1)\cdot a-(ax+b)\cdot (x^5+x+1)}{(x^5+x+1)^2}$
Now Equate  above equation, and get the value of $a$ and $b$
A: Let $$\displaystyle I = \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{\left[\left(5x^4+1\right)+(4x^5-1)\right]}{(x^5+x+1)^2}dx$$
So $$\displaystyle I = \underbrace{\int\frac{5x^4+1}{(x^5+x+1)^2}dx}_{J}+\underbrace{\int \frac{4x^5-1}{(x^5+x+1)^2}dx}_{K}$$
So for Calculation of 
$$\displaystyle J= \int \frac{5x^4+1}{(x^5+x+1)^2}dx$$
Let $(x^5+x+1) = t\;,$ Then $(5x^4+1)dx = dt$
So Integral $$\displaystyle J = \int\frac{1}{t^2}dt =-\frac{1}{t} = -\left(\frac{1}{x^5+x+1}\right)$$
Now for Calculation of 
$$\displaystyle K= \int\frac{4x^5-1}{(x^5+x+1)^2}dx = \int\frac{4x^5-1}{x^2\cdot \left(x^4+1+x^{-1}\right)^2}dx = \int \frac{\left(4x^3-x^{-2}\right)}{\left(x^4+1+x^{-1}\right)^2}dx$$
Now Put $\displaystyle \left(x^4+1+x^{-1}\right) = u\;,$ Then $\left(4x^3-x^{-2}\right)dx = du$
So Integral $$\displaystyle K = \int\frac{1}{u^2}du = -\frac{1}{u} = -\frac{1}{x^4+1+x^{-1}} = -\left(\frac{x}{x^5+x+1}\right)$$
So we get $$\displaystyle  I = J+K = -\frac{1}{x^5+x+1}-\frac{x}{x^5+x+1} = -\left(\frac{x+1}{x^5+x+1}\right)+\mathcal{C}$$
