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Question
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
Sum
Solution
\[\int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[\text{Let x - 1 }= t\]
\[ \Rightarrow x = t + 1\]
\[ \Rightarrow 1 = \frac{dt}{dx}\]
\[Now, \int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[ = \int\left[ \frac{2\left( t + 1 \right) - t}{t^2} \right]\text{ dt }\]
\[ = \int\left( \frac{2t + 1}{t^2} \right)\text{ dt }\]
\[ = 2\int\frac{dt}{t} + \int t^{- 2} \text{ dt }\]
\[ = \text{ 2 log }\left| t \right| + \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ =\text{ 2 log }\left( x - 1 \right) - \frac{1}{x - 1} + C\]
\[\text{Let x - 1 }= t\]
\[ \Rightarrow x = t + 1\]
\[ \Rightarrow 1 = \frac{dt}{dx}\]
\[Now, \int\left[ \frac{2x - 1}{\left( x - 1 \right)^2} \right]dx\]
\[ = \int\left[ \frac{2\left( t + 1 \right) - t}{t^2} \right]\text{ dt }\]
\[ = \int\left( \frac{2t + 1}{t^2} \right)\text{ dt }\]
\[ = 2\int\frac{dt}{t} + \int t^{- 2} \text{ dt }\]
\[ = \text{ 2 log }\left| t \right| + \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ =\text{ 2 log }\left( x - 1 \right) - \frac{1}{x - 1} + C\]
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