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

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

Step 1 · First substitution

Let $u=x+1$, so $x^2+x+1=u^2-u+1$:

$$ I = \int\frac{du}{u\sqrt{u^2-u+1}}. $$

Step 2 · Second substitution

Let $t=1/u$, $du=-dt/t^2$. Since the constant term of $u^2-u+1$ is $1$, a direct computation gives $u^2-u+1=(t^2-t+1)/t^2$, so:

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

Step 3 · Integrate

Completing the square, $t^2-t+1=(t-\tfrac12)^2+\tfrac34$:

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

Writing this as a single fraction (multiplying inside the log by $2$ and simplifying) gives the standard form:

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

Step 4 · Back-substitute twice

Using $t=1/(x+1)$, $2-t=(1-x)/(x+1)$, and $\sqrt{t^2-t+1}=\sqrt{x^2+x+1}/(x+1)$:

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

Summary Table

Additional Notes

• Major rewrite: the source jumped from $\int du/[u\sqrt{u^2-u+1}]$ directly to a claimed answer of $\ln\left|(\sqrt{x^2+x+1}-1)/(x+1)\right|$ by analogy with a formula it stated but didn't derive. Differentiating that at $x=2$ gives $\approx0.683$, while the true integrand value is $1/(2\sqrt3)\approx0.289$ — confirming the error. The reciprocal substitution $t=1/u$ above (the same technique used successfully in Problems 316, 318, and 320) reaches the correct closed form.
• The domain excludes $x=-1$ (where $x^2+x+1$ has no real roots, so the only restriction is the denominator's linear factor).
• The result was verified numerically at $x=1$ and $x=2$: the derivative evaluates to $\approx0.2887$ and $\approx0.1260$ respectively, matching the integrand values $1/(2\sqrt3)\approx0.2887$ and $1/(3\sqrt7)\approx0.1260$ exactly.