# Evaluating the integral $\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}}dx$

The integral to be evaluated is $$\int \frac{(x-1)}{(x+1) \sqrt{x^3+x^2+x}}dx$$

I split the integral to obtain $$\int \frac{\sqrt x}{(x+1) \sqrt{x^2+x+1}}dx - \int \frac{1}{(x+1)\sqrt{x^3+x^2+x}}dx$$

But I could not proceed any further, as I am not able to find any substitution, nor could I find any further simplification.

I tried by multiplying the numerator and denominator with $\sqrt {x-1}$ so as to obtain $x^3-1$ inside the radical, but that didn't help either.

How should I proceed to evaluate this integral?

• Wolfram|Alpha suggests that the indefinite integral can be expressed in terms of elliptic integrals of the first and third kind. – projectilemotion May 18 '17 at 16:30
• look like WA is not that smart. $2\tan^{-1}\sqrt{x+1+x^{-1}}$ – achille hui May 18 '17 at 16:52

Notice \begin{align} \int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}dx = &\int \frac{x-1}{(x+1)\sqrt{x+1+x^{-1}}}\frac{dx}{x}\\ = &\int \frac{x-1}{(x+1)\sqrt{x+1+x^{-1}}}\frac{d(x+1+x^{-1})}{x-x^{-1}}\\ = &\int \frac{d(x+1+x^{-1})}{(x+2+x^{-1})\sqrt{x+1+x^{-1}}} \end{align} Let $u = \sqrt{x + 1 +x^{-1}}$, the indefinite integral evaluates to $$\int \frac{du^2}{(1+u^2)u} = 2\int \frac{du}{1+u^2} = 2\tan^{-1}(u) + \text{ const.} = 2\tan^{-1}\sqrt{x+1+x^{-1}} + \text{ const.}$$