Advanced Integration techniques: Quadratic Expressions and U-Substitution Find $$\int  \frac{2x-1}{x^2-6x+13}dx $$
In the final steps after a u-substitution, one arrives at $$\int \frac{2u}{u^2+4}du + \int\frac{5}{ u^2+4}du$$
The next step is arriving at $$\ln(u^2+4) + 5\arctan(\frac{u}{2}) + C$$
How does $\int$ $dx(2u)\over(u^2+4)$ yield $\ln(u^2+4)$? The power of the denominator is two. The denominator is not equivalent to  a variable to the first power, so I do not think ln(...) can be the answer. 

Letting $u = x - 3$ we have that $du = dx$ and $2u + 5 = 2x -1$. 
$$\begin{align}\int \frac{2x- 1}{x^2-6x + 13}dx &= \int \frac{2x- 1}{(x-3)^2 + 4}dx\\&=\int \frac{2u + 5}{u^2 + 4}du\\&=\int \frac{2u}{u^2 + 4}du + \int \frac{5}{u^2 + 4}du\\&=\ln |u^2 + 4| + \frac{5}{2}\arctan\Big(\frac{u}{2}\Big) \end{align}$$
Because $\frac{1}{u^2 + 4} = \frac{1}{4}\frac{1}{\frac{u^2}{4}+1} = \frac{1}{2}\frac{1}{2}\frac{1}{\frac{u^2}{4}+1} = \frac{d}{du}\arctan (\frac{u}{2})$.
