Q.Evaluate the following.
$\int \frac{1}{4x^2 - 20x + 17}\,dx$

Complete the square: $4x^2 - 20x + 17 = 4\left[(x-\tfrac52)^2 - 2\right]$. Then apply $\int \frac{du}{u^2 - a^2} = \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + c$ with $u = x - \tfrac52$ and $a = \sqrt{2}$.

Step 1 — Complete the square in the denominator:

$4x^2 - 20x + 17 = 4\left(x^2 - 5x\right) + 17 = 4\left[\left(x - \tfrac{5}{2}\right)^2 - \tfrac{25}{4}\right] + 17$

$= 4\left(x - \tfrac{5}{2}\right)^2 - 25 + 17 = 4\left(x - \tfrac{5}{2}\right)^2 - 8 = 4\left[\left(x - \tfrac{5}{2}\right)^2 - 2\right]$

Step 2 — Rewrite the integral:

$\int \frac{dx}{4x^2 - 20x + 17} = \frac{1}{4}\int \frac{dx}{\left(x - \tfrac{5}{2}\right)^2 - \left(\sqrt{2}\right)^2}$

Step 3 — Apply the standard result $\int \frac{du}{u^2 - a^2} = \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + c$, with $u = x - \tfrac{5}{2}$ and $a = \sqrt{2}$:

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