Evaluate the integral: I = integral dx/(x sqrt(x²+x+1))

Evaluate the integral: $$ I = \int \frac{dx}{x\sqrt{x^2+x+1}}. $$

Step 1 · Reciprocal substitution

Let $x=1/t$, $dx=-dt/t^2$ (taking $t>0$, i.e. $x>0$). Then:

$$ x^2+x+1 = \frac{1+t+t^2}{t^2}, \qquad \sqrt{x^2+x+1} = \frac{\sqrt{t^2+t+1}}{t}. $$

Step 2 · Simplify

$$ I = \int \frac{-dt/t^2}{\dfrac1t\cdot\dfrac{\sqrt{t^2+t+1}}{t}} = -\int \frac{dt}{\sqrt{t^2+t+1}}. $$

Step 3 · Apply the standard logarithmic formula

$$ I = -\ln\left|t+\frac12+\sqrt{t^2+t+1}\right| + C. $$

Step 4 · Substitute back and combine

With $t=1/x$, multiplying the argument of the log by $2x$ and absorbing the resulting constant into $C$ gives the cleaner combined form:

$$ I = -\ln\left|2+x+2\sqrt{x^2+x+1}\right| + \ln|x| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$; the integrand is defined for all other real $x$, since $x^2+x+1>0$ everywhere.
• The reciprocal substitution $x=1/t$ is the standard trick whenever $x$ appears multiplied outside a radical of a quadratic — it moves the troublesome $x$ factor inside the radical, where it combines with the quadratic to produce another quadratic in $t$.
• The result was verified directly by differentiation (bypassing any sign subtleties from the $t$-substitution): writing $s=\sqrt{x^2+x+1}$, the derivative of $-\ln(2+x+2s)+\ln|x|$ simplifies, after combining over a common denominator, to numerator $x+2+2s$ over $xs(2+x+2s)$ — and since $x+2+2s$ is identical to the $(2+x+2s)$ factor in the denominator, these cancel, leaving exactly $1/(xs)=1/(x\sqrt{x^2+x+1})$, matching the integrand for all $x\neq0$.