Tedious rational integral

Question

I have been trying to solve this tedious integral, but I am unable to find a substitution that helps to solve it. $$\int\frac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}dx$$

Answer 1

There might be some things to attempt before substitution. First, we can factor some things out of the radicals.

$$\int\dfrac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}=\int\dfrac{\sqrt{x^2+x+2-2\sqrt{x^2+x+1}}}{x^2\sqrt{x^2+x+1}}=\int\dfrac{\sqrt{x^2+x+1-2\sqrt{x^2+x+1}+1}}{x^2\sqrt{x^2+x+1}}=\int\dfrac{\sqrt{x^2+x+1}-1}{x^2\sqrt{x^2+x+1}}$$

Can you find a suitable substitution now?

Answer 2

Starting with Mike’s wonderful simplification when $x>0$, we have $$ I=-\frac{1}{x}-\int \frac{d x}{x^2 \sqrt{x^2+x+1}} $$ Let $u=\frac 1x$ and then $$ \begin{aligned} \int \frac{1}{x^2 \sqrt{x^2+x+1}} d x = & \int \frac{1}{\frac{1}{u^2} \sqrt{\frac{1}{u^2}+\frac{1}{u}+1}} \frac{d u}{-u^2}\\ = & -\int \frac{u}{\sqrt{u^2+u+1}} d u \\ = & -\frac{1}{2} \int \frac{2 u+1-1}{\sqrt{u^2+u+1}} d u \\ = & -\sqrt{u^2+u+1+} \int \frac{d u}{\sqrt{(2 u+1)^2+3}} \\ = & -\sqrt{u^2+u+1}+\frac{1}{2} \operatorname{arcsinh}\left(\frac{2 u+1}{\sqrt{3}}\right)+c\\ = & -\frac{\sqrt{x^2+x+1}}{x}+\frac{1}{2} \operatorname{arcsinh}\left(\frac{x+2}{\sqrt{3} x}\right)+c \end{aligned} $$ Hence $$\boxed{I=-\frac{1}{x} +\frac{\sqrt{x^2+x+1}}{x}-\frac{1}{2} \operatorname{arcsinh}\left(\frac{x+2}{\sqrt{3} x}\right)+C}$$