evaluate $ \int \frac{4x^5-1}{(x^5+x+1)^2}dx$ As an exercise in section "Integration By Partial Fraction"  I tried to evaluate  $$ \int \frac{4x^5-1}{(x^5+x+1)^2}dx$$ by partial fraction,and I faild to solve it.After I see the answer it's evident that the derivative of $- \frac{x}{x^5+x+1}$ is $\frac{4x^5-1}{(x^5+x+1)^2}$,but I don't think it's evident if I didn't look at the answer.Is there any technique to evaluate integrals like this?
Thanks.
Progress:
I get an idea that if a rational function is in this form $$ \frac{P(x)}{G(x)^2} $$G(x) is a polynomial of degree $n$,and P(x) is a polynomial of degree $k( n-1 \leq k \leq2n-1)$,then we could suppose a polynomial  of degree $k-n+1$ $$f(x)= \sum_{i=0}^{k-n+1} c_i x^i$$ and use  the numerator part of quotient rule we can list an equation: $$f'(x)G(x)-f(x)G'(x)=P(x)$$ and  get $k+1$ linear equations,and use linear Algebra to check it whether it has a solution.
ATTENTION
This ONLY works for some particular partial fraction problem. It may be useful on an exam where the problems are intentionally simple. The general way to integrate rational function is using partial fraction.
 A: The quotient rule may provide a way to reverse engineer the problem. (the denominator being of the form $G(x)^2$)
Here goes... $$d(f(x)(x^5+x+1)^{-1})= [f'(x)(x^5+x+1)-(5x^4+1)(f(x)]\space(\text {over $g(x)^2$})$$ Then $$f'(x)(x^5+x+1)+(-5x^4-1)f(x)=4x^5-1$$ since I need to have only terms of $x^5$ while I have $(-5x^4-1)$ f(x) and $f'(x)$ must not raise the power of the $x^5$ term I already have, f(x)must be $\frac{+}{-}x$ and since I need $+4x^5,f(x)\text{must be} -x$
This will yield the true statement
 $$-x^5-x-1+5x^5+x=4x^5-1$$ So the integral must be $$-x(x^5+x+1)^{-1}$$All that being said, ( you wanted an easier way in THIS case) this ONLY works because the denominator is of the form $g(x)^2$, an even power, (and it will get unbearably hairy for any even power over 2); and the numerator is so clean. 
I am sure these folks will tell you ( and I am inclined to agree) that it is better to learn Integration by partial fractions.
A: Here is an actual method to solve this integral, using integration by parts.
$\begin{align}
I&=\int\frac{4x^5-1}{(x^5+x+1)^2}\,\mathrm{d}x\\
&=\int\frac{4x^5-1}{5x^4+1}\frac{5x^4+1}{(x^5+x+1)^2}\,\mathrm{d}x\\
&=\frac{1-4x^5}{(5x^4+1)(x^5+x+1)}+\int\frac{20x^3}{(5x^4+1)^2}\,\mathrm{d}x\\
&=\frac{-x}{x^5+x+1}+C
\end{align}$
A: Actually, there is a way to solve this integral only with substitution. Note that
$$\begin{aligned}
\int \frac{4x^5 - 1}{\left(x^5 + x + 1\right)^2}\,\mathrm{d}x&=\int\frac{4x^5 - 1}{\left[x\left(x^4 + 1/ x + 1\right)\right]^2}\,\mathrm{d}x\\
&=\int\frac{4x^3 - 1/x^2}{\left(x^4 + 1/x + 1\right)^2}\,\mathrm{d}x
\end{aligned} $$
Now, set $u = x^4 + 1/x+1 $ and $\mathrm{d}u = \left(4x^3 - 1/x^2\right)\mathrm{d}x$ to obtain
$$\int\frac{\mathrm{d}u}{u^2} = -\frac{1}{u} + C. $$
Then,
$$\begin{aligned}
\int \frac{4x^5 - 1}{\left(x^5+x+1\right)^2}\,\mathrm{d}x &= -\frac{1}{x^4 + 1/x+1}+C\\
&=-\frac{x}{x^5 +x+1} + C,
\end{aligned}$$
as desired.
