Evaluate the integral: I = integral dx/(sqrt(2x²-6x+5))

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{2x^2-6x+5}}. $$

Step 1 · Complete the square

$$ 2x^2-6x+5 = 2\left(x-\frac32\right)^2 + \frac12. $$

So:

$$ \sqrt{2x^2-6x+5} = \sqrt2\,\sqrt{\left(x-\frac32\right)^2+\frac14}. $$

Step 2 · Substitute

Let $u=x-\tfrac32$, $du=dx$, $a=\tfrac12$:

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

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 $u=x-\tfrac32$ and $u^2+a^2=x^2-3x+\tfrac94+\tfrac14=x^2-3x+\tfrac52=\tfrac12(2x^2-6x+5)$, so $\sqrt{u^2+a^2}=\sqrt{2x^2-6x+5}/\sqrt2$ — not simply $\sqrt{2x^2-6x+5}$:

$$ I = \frac{1}{\sqrt2}\ln\left|x-\frac32+\frac{\sqrt{2x^2-6x+5}}{\sqrt2}\right| + C. $$

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

$$ I = \frac{1}{\sqrt2}\ln\left|2x-3+\sqrt2\sqrt{2x^2-6x+5}\right| + C. $$

Summary Table

Additional Notes

• Correction: the source wrote $\sqrt{(x-3/2)^2+1/4}=\sqrt{2x^2-6x+5}$ directly, but this drops a factor of $\sqrt2$ — the correct relation is $\sqrt{(x-3/2)^2+1/4}=\sqrt{2x^2-6x+5}/\sqrt2$. Differentiating the source's stated final answer at $x=3/2$ gives $1$, but the true derivative there should be $1/\sqrt{0.5}=\sqrt2\approx1.414$, confirming the error.
• The integrand is defined for all real $x$, since $2x^2-6x+5>0$ everywhere (discriminant $36-40=-4$).
• The result was verified numerically: differentiating $\tfrac{1}{\sqrt2}\ln|2x-3+\sqrt2\sqrt{2x^2-6x+5}|$ at $x=1.5$ gives exactly $\sqrt2$, matching $1/\sqrt{2(1.5)^2-6(1.5)+5}=1/\sqrt{0.5}=\sqrt2$.