Question
Find the integral: $\int \frac{d x}{x^{2}-6 x+13}$

Answer

We have $x^2 - 6x + 13 $
$= x^2 - 6x + 3^2 - 3^2 + 13 $
$= (x - 3)^2 + 4$
So, $\int \frac{d x}{x^{2}-6 x+13}$
$=\int \frac{1}{(x-3)^{2}+2^{2}} d x$
Let $x – 3 = t$
$\Rightarrow dx = dt$
Therefore, $\int \frac{d x}{x^{2}-6 x+13}$
$=\int \frac{d t}{t^{2}+2^{2}}$
$=\frac{1}{2} \tan ^{-1} \frac{t}{2}+\mathrm{C}$
$=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+C$

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