There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ (ln(1 - x) + x){\frac{1}{x}}^{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{ln(-x + 1)}{x^{2}} + \frac{1}{x}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{ln(-x + 1)}{x^{2}} + \frac{1}{x}\right)}{dx}\\=&\frac{-2ln(-x + 1)}{x^{3}} + \frac{(-1 + 0)}{x^{2}(-x + 1)} + \frac{-1}{x^{2}}\\=&\frac{-2ln(-x + 1)}{x^{3}} - \frac{1}{(-x + 1)x^{2}} - \frac{1}{x^{2}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2ln(-x + 1)}{x^{3}} - \frac{1}{(-x + 1)x^{2}} - \frac{1}{x^{2}}\right)}{dx}\\=&\frac{-2*-3ln(-x + 1)}{x^{4}} - \frac{2(-1 + 0)}{x^{3}(-x + 1)} - \frac{(\frac{-(-1 + 0)}{(-x + 1)^{2}})}{x^{2}} - \frac{-2}{(-x + 1)x^{3}} - \frac{-2}{x^{3}}\\=&\frac{6ln(-x + 1)}{x^{4}} + \frac{4}{(-x + 1)x^{3}} - \frac{1}{(-x + 1)^{2}x^{2}} + \frac{2}{x^{3}}\\ \end{split}\end{equation} \]

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