Evaluate the integral: I = integral sqrt(2+x+x²) dx

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

Explanation of Method

Completing the square converts the radicand into the standard $u^2+a^2$ form, resolved by the well-known formula $\int\sqrt{u^2+a^2}\,du=\tfrac u2\sqrt{u^2+a^2}+\tfrac{a^2}2\ln|u+\sqrt{u^2+a^2}|$.

Complete the square for $2+x+x^2$.
Substitute $u=x+\tfrac12$.
Apply the standard formula for $\int\sqrt{u^2+a^2}\,du$.
Substitute back.

Step 1 · Complete the square

$$ 2+x+x^2 = \left(x+\frac12\right)^2 + \frac74. $$

Step 2 · Substitute

Let $u=x+\tfrac12$, $du=dx$, $a^2=\tfrac74$:

$$ I = \int \sqrt{u^2+\frac74}\,du. $$

Step 3 · Apply the standard formula

$$ I = \frac u2\sqrt{u^2+\frac74} + \frac78\ln\left|u+\sqrt{u^2+\frac74}\right| + C. $$

Step 4 · Substitute back

With $u=x+\tfrac12$, so $\tfrac u2=\tfrac{2x+1}4$:

$$ I = \frac{2x+1}{4}\sqrt{2+x+x^2} + \frac78\ln\left|x+\frac12+\sqrt{2+x+x^2}\right| + C. $$

Summary Table

Quantity	Value
Completed square	$(x+\tfrac12)^2+\tfrac74$
$u,\,a^2$	$x+\tfrac12,\ \tfrac74$

Additional Notes

Correction: the source's final answer had two separate slips: the linear coefficient was written as $(2x-1)/4$ instead of $(2x+1)/4$ (sign error, since $u/2=(x+1/2)/2=(2x+1)/4$), and the log coefficient was written as $7/2$ instead of $7/8$ (since $a^2/2=(7/4)/2=7/8$, not $7/2$).
The integrand is defined for all real $x$, since $2+x+x^2>0$ everywhere (discriminant $1-8=-7$).
The result was verified by differentiation: writing the product-rule and log-rule derivatives of $\tfrac{2x+1}4\sqrt{2+x+x^2}+\tfrac78\ln|x+\tfrac12+\sqrt{2+x+x^2}|$ over the common denominator $8\sqrt{2+x+x^2}$ gives numerator $4(2+x+x^2)+(2x+1)^2+7=8x^2+8x+16=8(x^2+x+2)$, which reduces exactly to $\sqrt{2+x+x^2}$.

Result

The solution to the integral is:

$$ I = \frac{2x+1}{4}\sqrt{2+x+x^2} + \frac78\ln\left|x+\frac12+\sqrt{2+x+x^2}\right| + C. $$