Evaluation of $\int\frac{1}{x+ \sqrt{x^2-x+1}}dx$

Question

Evaluate :

$$\int\frac{1}{x+ \sqrt{x^2-x+1}}dx$$

After multiplying the denominator with $x-\sqrt{x^2-x+1}$, I get

$$x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx$$

Is $\int \frac{\sqrt{x^2-x+1}}{x-1}dx$ is integrable in terms of elementary functions?

Answer 1

We have $$I =\int \frac {\sqrt {x^2-x+1}}{x-1} dx =\frac {1}{2 }\int \frac {\sqrt {(2x-1)^2+3}}{x-1} dx $$ Substituting $u =2x-1$, we get, $$I =\frac {1}{2} \int \frac {\sqrt{u^2+3}}{u-1} du $$ Now perform hyperbolic substitution $u =\sqrt {3} \operatorname {sinh}(v)$ and simplifying gives us $$I =\frac {3}{2} \int \frac {(\operatorname {cosh }(v))^2}{\sqrt {3}\operatorname{sinh}(v) -1} dv $$ which can be solved by Weierstrass substitution. Hope it helps.

Answer 2

There is a substitution, called the Euler substitution, which solves your integral rather neatly. In your case we set $t = \sqrt{x^2 - x + 1} + x$; after subtracting $x$ from both sides and squaring, we get $x^2 -x + 1 = t^2 - 2xt + x^2$, which can be solved (after subtracting $x^2$ from both sides) to get $x = \frac{t^2 - 1}{2t - 1}$. In this case, $\frac{dt}{dx} = \frac{1}{\sqrt{x^2 - x + 1}}(2x-1) + 1$, and so

\begin{eqnarray} \int \frac{dx}{x + \sqrt{x^2 - x + 1}} = \int (\frac{1}{x + \sqrt{x^2 - x + 1}}\cdot\frac{1}{\frac{dt}{dx}})\frac{dt}{dx}dx \\= \int \frac{1}{t}\cdot\frac{1}{[\frac{1}{t - (t^2-1)/(2t-1)}\cdot(2(\frac{t^2-1}{2t-1}) - 1)] + 1} dt = \dots \\ \dots = \int\frac{t^2 - t + 1}{3t^3 - 3t^2} dt\end{eqnarray}

which is quite easy to solve directly by partial fractions.

Answer 3

The modified Euler sub $t=\sqrt{x^2-x+1}-x+\frac12$ is best suited here

\begin{align} &\int\frac{1}{x+ \sqrt{x^2-x+1}}\ dx\\ =& -\int \frac{3+4t^2}{2t(3+2t)}\ dt=-t-\frac12\ln t+2 \ln(2t+3)+C \end{align}