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

Step 1 ：Rewrite the given function using partial fraction decomposition: $$\frac{10}{(x-1)(x^2+9)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+9}$$

Step 2 ：Find the values of A, B, and C: $$A = 1, B = -1, C = -1$$

Step 3 ：Rewrite the function with the found values: $$\frac{10}{(x-1)(x^2+9)} = \frac{1}{x-1} - \frac{x+1}{x^2+9}$$

Step 4 ：Integrate each term separately: $$\int \frac{1}{x-1} dx$$ and $$\int \frac{-x-1}{x^2+9} dx$$

Step 5 ：Find the integrals: $$\int \frac{1}{x-1} dx = \log(x-1)$$ and $$\int \frac{-x-1}{x^2+9} dx = -\frac{1}{2}\log(x^2+9) - \frac{1}{3}\arctan\left(\frac{x}{3}\right)$$

Step 6 ：Combine the integrals to find the final answer: \(\boxed{\int \frac{10}{(x-1)\left(x^{2}+9\right)} d x = \log(x-1) - \frac{1}{2}\log(x^2+9) - \frac{1}{3}\arctan\left(\frac{x}{3}\right) + C}\)