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

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

Step 1 · Rewrite and split

$$ \frac{x^2}{\sqrt{1+x+x^2}} = \sqrt{1+x+x^2} - \frac{1+x}{\sqrt{1+x+x^2}}. $$

Step 2 · Evaluate the second piece

Writing $1+x=\tfrac12(1+2x)+\tfrac12$:

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

Step 3 · Evaluate the first piece

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

$$ \int\sqrt{1+x+x^2}\,dx = \frac{2x+1}4\sqrt{1+x+x^2} + \frac38\ln\left|x+\frac12+\sqrt{1+x+x^2}\right|. $$

Step 4 · Combine

$$ I = \left[\frac{2x+1}4-1\right]\sqrt{1+x+x^2} + \left[\frac38-\frac12\right]\ln\left|x+\frac12+\sqrt{1+x+x^2}\right| + C. $$

Summary Table

Additional Notes

• Correction: the source's Step 3 contained an algebra slip, writing "$1+x=\tfrac12+x$" (these are not equal — they differ by $\tfrac12$) before correctly factoring out $\tfrac12(1+2x)$. This dropped a constant term and produced a logarithm coefficient of $+\tfrac38$ in the final answer instead of the correct $-\tfrac18$. Differentiating the source's answer at $x=0$ gives $1.25$, while the true integrand there is $0$, confirming the error.
• The domain is all real $x$, since $1+x+x^2>0$ everywhere.
• The result was verified by differentiation: combining $\tfrac{d}{dx}\left[\tfrac{2x-3}4\sqrt{1+x+x^2}\right]$ and $\tfrac{d}{dx}\left[-\tfrac18\ln|x+\tfrac12+\sqrt{1+x+x^2}|\right]$ over the common denominator $2\sqrt{1+x+x^2}$ gives numerator $x^2$, matching the integrand.