Evaluate the indefinite algebraic integral $\int\frac{dx}{(x-1)\sqrt{x^2+x+1}}$

Question

I am trying to calculate $$I=\int\dfrac{dx}{(x-1)\sqrt{x^2+x+1}}$$

In my textbook, they advise us to put $x-1=\dfrac{1}{t}$, but I don't see how this helps. We will then have $$I=\int\dfrac{d\frac{1}{t}}{\frac{1}{t}\sqrt{(1+\frac{1}{t})^2+1+\frac{1}{t}+1}}=\int\dfrac{d\frac{1}{t}}{\frac{1}{t}\sqrt{(\frac{1}{t}+\frac32)^2+\frac34}}$$

How is this useful? Thanks!

Answer 1

Simplifying gives (assuming $t > 0$, hence $x > 1$) that $$I = -\int \frac{dt}{\sqrt{3 t^2 + 3 t + 1}} = - \frac{1}{\sqrt{3}} \int \frac{du}{\sqrt{1 + u^2}}.$$ (If $t < 0$, we simply pick up an additional $-$ sign.)

The second equality follows from substituting $t = \frac12\left(\frac1{\sqrt 3} u - 1\right)$.

Another option is the apply the Euler Substitution of the First Kind, $\sqrt{x^2 + x + 1} = x + s$, to the original integral in $x$, which rationalizes it, after which the affine transformation $s = \sqrt 3 y - 1$ puts it in a standard form. $$I = 2 \int \frac{ds}{s^2 + 2 s - 2} = -\frac2{\sqrt 3} \int \frac{dy}{1 - y^2} .$$ We can then proceed as usual.

Answer 2

To avoid worrying about signs pulling the variable out of square-root, integrate instead with $$g(t)=\frac{ 3(1+\frac12 t)}{\sqrt{t^2+3t+3}}\>\implies g’(t)= \frac{g^2(t)-3}{t \sqrt{t^2+3t+3}} $$ Then $$\int \frac1{t \sqrt{t^2+3t+3}}dt= \int\frac{g’(t)}{g^2(t)-3}dt=-\frac1{\sqrt3}\tanh^{-1}\frac{g(t)}{\sqrt3}+C $$ which is valid for all domain $t$.

Answer 3

To get rid of $(x-1)$, we let $y=\frac{1}{x-1}$ and get $$ I=\int \frac{d x}{(x-1) \sqrt{x^2+x+1}}=-\int \frac{d y}{\sqrt{3 y^2+3 y+1}}= -\frac{2}{\sqrt{3}} \int \frac{d y}{\sqrt{(2 y+1)^2+\frac{1}{3}}} $$ Letting $2 y+1=\frac{1}{\sqrt{3}} \sinh \theta$ yields $$ \begin{aligned} I & =-\frac{\sinh ^{-1}(\sqrt{3}(2 y+1))}{\sqrt{3}}+C \\ & =-\frac{1}{\sqrt{3}} \sinh ^{-1}\left(\frac{\sqrt{3}(x+1)}{x-1}\right)+C \end{aligned} $$