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. 
 A: Probably they missed include a constant in book's answer.
If we include a constant $k$, the book's answer will change to:
$$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k$$
But with some algebra we get
$$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k=$$ $$=5(\frac{(x-5)}{x-5}\ln|x-5|-\frac{x}{x-5})+k= 5(\ln|x-5|-\frac{x}{x-5}+1)-5+k=$$ $$=5(\ln|x-5|-\frac{x}{x-5}+\frac{x-5}{x-5})-5+k=5(\ln|x-5|-\frac{5}{x-5})-5+k=$$
$$=5(\ln|x-5|-\frac{5}{x-5})+C$$
Which is your answer, where C is a new constant, such that $C=k-5$.
A: 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:
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x= \frac{5}{2}\int\frac{2x-10}{\left(x-5\right)^2}+\frac{10}{\left(x-5\right)^2}\,\mathrm{d}x=\frac{5}{2}\left(2\ln\vert x-5\vert-\frac{10}{x-5}\right)+\mathcal{C}$$
A: Distribute:
$$
\frac{5}{x-5}((x-5)\ln|x-5|-x)=5\left(\frac{x-5}{x-5}\ln|x-5|-\frac{x}{x-5}\right).
$$
Then
$$
\frac{x-5}{x-5}=1.
$$
