Jtdylktuy

Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$

The book hint is:

Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?

The book answer is:

• Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
  

It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.