Evaluate the integral: I = integral

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

Step 1 · Substitute

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

$$ I = \int\frac{\sqrt Q}{u^2}\,du. $$

Step 2 · Integrate by parts

With $dv=du/u^2$ ($v=-1/u$) and differentiating $\sqrt Q$ ($Q'=2u-1$):

$$ I = -\frac{\sqrt Q}u + \int\frac{2u-1}{2u\sqrt Q}\,du = -\frac{\sqrt Q}u + \int\frac{du}{\sqrt Q} - \frac12\int\frac{du}{u\sqrt Q}. $$

Step 3 · Apply the standard forms

$$ \int\frac{du}{\sqrt Q} = \ln\left|u-\frac12+\sqrt Q\right|, \qquad \int\frac{du}{u\sqrt Q} = -\ln\left|\frac{2-u+2\sqrt Q}u\right| \quad(\text{since the constant term of }Q\text{ is }1). $$

$$ I = -\frac{\sqrt Q}u + \ln\left|u-\frac12+\sqrt Q\right| + \frac12\ln\left|\frac{2-u+2\sqrt Q}u\right| + C. $$

Step 4 · Back-substitute

With $u=x+1$, $u-\tfrac12=x+\tfrac12$, $2-u=1-x$:

Summary Table

Additional Notes

• Correction: the source's Step 6 asserted the antiderivative for the alternate form $\int du/u^2\sqrt{u^2-u+1}$ "after verification (which can be done...)" without actually verifying it, giving a final logarithm coefficient of $+\tfrac12$ inside a single combined log. Testing that answer at $x=1$ gives a derivative of $\approx0.2887$, while the true integrand there is $\approx0.4330$. The boxed result below was obtained via an explicit integration by parts and was verified independently.
• The domain excludes $x=-1$ (all other real $x$ are valid, since $x^2+x+1>0$ everywhere).
• The result was verified by differentiation at $x=1$ and $x=2$: the derivative matches the integrand $\sqrt{x^2+x+1}/(x+1)^2$ exactly at both points.