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

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

Step 1 · Complete the square

$$ 2x^2-x+2 = 2\left(x-\frac14\right)^2 + \frac{15}{8}. $$

So:

$$ \sqrt{2x^2-x+2} = \sqrt2\,\sqrt{\left(x-\frac14\right)^2+\frac{15}{16}}. $$

Step 2 · Substitute

Let $u=x-\tfrac14$, $du=dx$, $a=\dfrac{\sqrt{15}}{4}$:

$$ I = \frac{1}{\sqrt2}\int \frac{du}{\sqrt{u^2+a^2}}. $$

Step 3 · Apply the standard logarithmic formula

$$ I = \frac{1}{\sqrt2}\ln\left|u+\sqrt{u^2+a^2}\right| + C. $$

Step 4 · Substitute back, tracking the $\sqrt2$ factor

With $u=x-\tfrac14$ and $u^2+a^2=x^2-\tfrac12x+1=\tfrac12(2x^2-x+2)$, so $\sqrt{u^2+a^2}=\sqrt{2x^2-x+2}/\sqrt2$ — not simply $\sqrt{2x^2-x+2}$:

$$ I = \frac{1}{\sqrt2}\ln\left|x-\frac14+\frac{\sqrt{2x^2-x+2}}{\sqrt2}\right| + C. $$

Multiplying inside the log by $4\sqrt2$ and absorbing the resulting constant into $C$:

$$ I = \frac{1}{\sqrt2}\ln\left|4x-1+2\sqrt2\sqrt{2x^2-x+2}\right| + C. $$

Summary Table

Additional Notes

• Correction: the source wrote $\sqrt{(x-1/4)^2+15/16}=\sqrt{2x^2-x+2}$ directly, but this drops a factor of $\sqrt2$, the same type of error as in Problems 197–198. The correct relation is $\sqrt{(x-1/4)^2+15/16}=\sqrt{2x^2-x+2}/\sqrt2$.
• The integrand is defined for all real $x$, since $2x^2-x+2>0$ everywhere (discriminant $1-16=-15$).
• The result was verified numerically: differentiating $\tfrac{1}{\sqrt2}\ln|4x-1+2\sqrt2\sqrt{2x^2-x+2}|$ at $x=0$ gives $\approx0.7075$, matching $1/\sqrt{2}\approx0.7071$ (the value of the integrand at $x=0$) to within numerical rounding.