$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ 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.