If ∫ d x x 2 - 2 x + 10 2 = A tan - 1 x - 1 3 + f x x 2 - 2 x + 10 + C , then (where C is a constant…

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

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

$\Rightarrow I=\int \frac{3\mathrm{s}\mathrm{e}{\mathrm{c}}^2\theta d\theta }{81\mathrm{s}\mathrm{e}{\mathrm{c}}^4\theta }=\frac{1}{27}\int \mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta d\theta$

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

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

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

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

$=\frac{1}{54}\left[\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\left(\frac{x-1}{3}\right)+\frac{3\left(x-1\right)}{x^2-2x+10}\right]+C$

Hence, $A=\frac{1}{54} \& f\left(x\right)=3(x-1)$