$ \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.