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

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

Step 1 · Rewrite the numerator

Since $\dfrac{d}{dx}(x^2+x+1)=2x+1$, write:

$$ x+1 = \frac12(2x+1) + \frac12. $$

Step 2 · Split the integral

$$ I = \frac12\int \frac{2x+1}{\sqrt{x^2+x+1}}\,dx + \frac12\int \frac{dx}{\sqrt{x^2+x+1}}. $$

Step 3 · Integrate the first piece

Let $u=x^2+x+1$, $du=(2x+1)\,dx$:

$$ \frac12\int \frac{2x+1}{\sqrt{x^2+x+1}}\,dx = \frac12\int \frac{du}{\sqrt u} = \sqrt u = \sqrt{x^2+x+1}. $$

Step 4 · Integrate the second piece

Completing the square, $x^2+x+1=(x+\tfrac12)^2+\tfrac34$, so using the standard logarithmic formula:

$$ \int \frac{dx}{\sqrt{x^2+x+1}} = \ln\left|x+\frac12+\sqrt{x^2+x+1}\right| + C. $$

Therefore:

$$ \frac12\int \frac{dx}{\sqrt{x^2+x+1}} = \frac12\ln\left|x+\frac12+\sqrt{x^2+x+1}\right|. $$

Step 5 · Combine

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

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $x^2+x+1>0$ everywhere.
• Unlike Problem 197's family, no leading-coefficient tracking is needed here when substituting back into the log term, since $x^2+x+1$ already has leading coefficient $1$.
• The result can be checked by differentiating $\sqrt{x^2+x+1}+\tfrac12\ln|x+\tfrac12+\sqrt{x^2+x+1}|$: the first term gives $\tfrac{2x+1}{2\sqrt{x^2+x+1}}$, and the second gives $\tfrac{1}{2\sqrt{x^2+x+1}}$ (after the $(\sqrt{x^2+x+1}+x+\tfrac12)$ factors cancel in the log's derivative); summing gives $\tfrac{2x+2}{2\sqrt{x^2+x+1}}=\tfrac{x+1}{\sqrt{x^2+x+1}}$, matching the integrand.