Integral by partial fractions 
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)

factor out the 5, and use partial fraction 
$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$
Solve for $A$ and $B$. 
$A\left(x-5\right) + B = x$  Then $B-5A$ has to be zero and $A$ has to be 1. 
Resulting in 
$$ 5 \left[\int \frac{1}{x-5} + \frac{5}{\left(x-5\right)^2}\, \mathrm{d}x \right]$$
$$ \Rightarrow 5 \left[ \ln \vert x - 5 \vert -\frac{5}{x-5}\right] + C$$
However, this approach doesn't give the answer in the book. 
Book's Answer 
$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x  \right) + C $$
The value should be 30, according to the book. 

Your solution is correct, but books solution is also. Differentiate the solutions and you will see, that both of them are Antiderivatives.
Moreover it is:
$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x  \right)
= 5 \left(\ln \vert x - 5 \vert - \frac{x-5+5}{x-5}\right) = 5 \left(\ln \vert x - 5 \vert - \frac{5}{x-5}\right) +\mathcal{Const}$$
Optional way to get your solution:
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x= \frac{5}{2}\int\frac{2x-10}{\left(x-5\right)^2}+\frac{10}{\left(x-5\right)^2}\,\mathrm{d}x=\frac{5}{2}\left(2\ln\vert x-5\vert-\frac{10}{x-5}\right)+\mathcal{C}$$
