constant of integration, then :

Let $I=\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=\int \frac{d x}{\left((x-1)^{2}+9\right)^{2}}$

Let $(x-1)^{2}=9 \tan ^{2} \theta$ ....(1)

$\Rightarrow \tan \theta=\frac{x-1}{3}$

After differentiating equation (1), we get

$2(x-1) d x=18 \tan \theta \sec ^{2} \theta d \theta$

$\therefore I=\int \frac{18 \tan \theta \sec ^{2} \theta d \theta}{2 \times 3 \tan \theta \times 81 \sec ^{4} \theta}$

$I=\frac{1}{27} \int \cos ^{2} \theta d \theta=\frac{1}{27} \times \frac{1}{2} \int(1+\cos 2 \theta) d \theta$

$I=\frac{1}{54}\left\{\theta+\frac{\sin 2 \theta}{2}\right\}+c$

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

$I=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{3(x-1)}{x^{2}-2 x+10}\right]+c$

Compare it with A $\left[\tan ^{-1}\left(\frac{x-1}{b}\right)+\frac{f(x)}{x^{2}-2 x+10}\right]+c$,

we get: $A=\frac{1}{54}$ and $f(x)=3(x-1)$