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Question
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Solution
\[\int\frac{2x}{\left( 2x + 1 \right)^2}dx\]
\[ = \int\left( \frac{2x + 1 - 1}{\left( 2x + 1 \right)^2} \right)dx\]
\[ = \int\left[ \frac{2x + 1}{\left( 2x + 1 \right)^2} - \frac{1}{\left( 2x + 1 \right)^2} \right]dx\]
\[ = \int\frac{dx}{2x + 1} - \int \left( 2x + 1 \right)^{- 2} dx\]
\[ = \frac{\log\left( 2x + 1 \right)}{2} - \left[ \frac{\left( 2x + 1 \right)^{- 2 + 1}}{2\left( - 2 + 1 \right)} \right] + C\]
\[ = \frac{\log \left( 2x + 1 \right)}{2} + \frac{\left( 2x + 1 \right)^{- 1}}{2} + C\]
\[ = \frac{\log \left( 2x + 1 \right)}{2} + \frac{1}{2\left( 2x + 1 \right)} + C\]
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