Evaluate the integral: I = integral dx/(x sqrt((1+x+x²)³))

Evaluate the integral: $$ I = \int \frac{dx}{x\sqrt{\left(1+x+x^2\right)^3}}. $$

Step 1 · Substitute

Let $u=1/x$, so $x=1/u$, $dx=-du/u^2$, and $1+x+x^2=1+\tfrac1u+\tfrac1{u^2}=\dfrac{u^2+u+1}{u^2}$, so $\left(1+x+x^2\right)^{3/2}=\dfrac{(u^2+u+1)^{3/2}}{|u|^3}$:

$$ I = \int\frac{-du/u^2}{\tfrac1u\cdot\dfrac{(u^2+u+1)^{3/2}}{u^3}} = -\int\frac{du}{\sqrt{(u^2+u+1)^3}}. $$

Step 2 · Complete the square

With $v=u+\tfrac12$, $u^2+u+1=v^2+\tfrac34$:

$$ I = -\int\frac{dv}{\left(v^2+3/4\right)^{3/2}}. $$

Step 3 · Apply the standard reduction

Using $\displaystyle\int\frac{dv}{\left(v^2+a^2\right)^{3/2}} = \frac v{a^2\sqrt{v^2+a^2}}$ with $a^2=\tfrac34$:

$$ I = -\frac{v}{\tfrac34\sqrt{v^2+3/4}} + C = -\frac{4v}{3\sqrt{v^2+3/4}} + C. $$

Step 4 · Back-substitute

With $v=u+\tfrac12=\tfrac1x+\tfrac12=\dfrac{2+x}{2x}$ and $\sqrt{v^2+3/4}=\sqrt{u^2+u+1}=\dfrac{\sqrt{1+x+x^2}}{|x|}$:

$$ I = -\frac43\cdot\frac{(2+x)/(2x)}{\sqrt{1+x+x^2}/x} + C = -\frac{2(2+x)}{3\sqrt{1+x+x^2}} + C. $$

Summary Table

Additional Notes

• Major rewrite: the source's proposed answer had the wrong structural form entirely (it included spurious logarithmic terms not present for this rational-type integrand). Since $\int dx/(x\sqrt{quadratic^3})$ reduces — after $u=1/x$ — to the purely algebraic standard form $\int dv/(v^2+a^2)^{3/2}$, no logarithm should appear at all. The boxed result below was obtained by a full independent rederivation via the reciprocal substitution.
• The domain excludes $x=0$ (all other real $x$ are valid, since $1+x+x^2>0$ everywhere).
• The result was verified by differentiation at $x=1$: the boxed expression's derivative evaluates to $1/(3\sqrt3)\approx0.19245$, matching the integrand $1/\left[x\sqrt{(1+x+x^2)^3}\right]=1/\sqrt{27}\approx0.19245$ at $x=1$ exactly.