If ∫ dx ( x ^ 2 - 2x + 10) ^ 2 = A( ^ - 1 ((x - 1)/(3)) + f(x) x ^...

We have $I = ∫ \dfrac{dx}{{\left({x}^{2} - 2x + 10\right)}^{2}} = ∫ \dfrac{dx}{{\left[{\left(x - 1\right)}^{2} + {3}^{2}\right]}^{2}}$

Let $x – 1 = 3tan\theta ⇒ dx = 3\sec^{2}\thetad\theta$

$⇒ I = ∫ \dfrac{3se{c}^{2}\thetad\theta}{81se{c}^{4}\theta} = \dfrac{1}{27} ∫ co{s}^{2}\thetad\theta$

$= \dfrac{1}{54} ∫ \left(1 + \cos 2\theta\right)d\theta$

$= \dfrac{1}{54}\left(\theta + \dfrac{\sin 2\theta}{2}\right) + C$

$= \dfrac{1}{54}\left(\theta + sin\theta cos\theta\right) + C$, where $C$ is the constant of integration.

$= \dfrac{1}{54}\left[\tan^{- 1}\left(\dfrac{x - 1}{3}\right) + \dfrac{\left(\dfrac{x - 1}{3}\right)}{1 + {\left(\dfrac{x - 1}{3}\right)}^{2}}\right] + C$

$= \dfrac{1}{54}\left[ta{n}^{- 1}\left(\dfrac{x - 1}{3}\right) + \dfrac{3\left(x - 1\right)}{{x}^{2} - 2x + 10}\right] + C$

Hence, $A = \dfrac{1}{54} \text{ and } f\left(x\right) = 3 ( x - 1 )$