I am following a worked example in the textbook Mathematical Methods for Physics and Engineering. Below is the part of the problem that I am having difficulty with. After manipulating $ I $ to give the first line, logarithmic integration is used to compete the problem. On the penultimate line on the first term there is a negative sign. What is its origin?
$$ \begin{align} I &= \int \frac{1}{\sqrt{2}} ( \frac{1}{\sqrt{2}-t} + \frac{1}{\sqrt{2} + t} ) dt \\ &= - \frac{1}{\sqrt{2}} \ln (\sqrt{2} - t) + \frac{1}{\sqrt{2}} \ln(\sqrt{2} + t) + c \\ &= \frac{1}{\sqrt{2}} \ln \frac{\sqrt{2} + \tan(\frac{x}{2})}{\sqrt{2} - \tan(\frac{x}{2})} + c \end{align} $$
The chain rule applied to the derivative of $$f(x) = \log(1-x)$$ implies $$\frac{df}{dx} = \frac{1}{1-x} \cdot \frac{d}{dx}[1-x] = -\frac{1}{1-x},$$ so the antiderivative $$\int \frac{1}{1-x} \, dx = -\log (1-x) + C.$$