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Question
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Solution
\[\int\frac{x^2}{\left( x - 1 \right)^3}\text{ dx }\]
\[ = \int\left[ \frac{x^2 - 1 + 1}{\left( x - 1 \right)^3} \right]\text{ dx }\]
\[ = \int\left[ \frac{\left( x - 1 \right) \left( x + 1 \right)}{\left( x - 1 \right)^3} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]
\[ = \int\left[ \frac{x + 1}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]
\[ = \int\left[ \frac{x - 1 + 2}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]
\[ = \int\left[ \frac{1}{\left( x - 1 \right)} + \frac{2}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]
\[ = \int\frac{1}{x - 1}\text{ dx }+ 2\int \left( x - 1 \right)^{- 2} \text{ dx }+ \int \left( x - 1 \right)^{- 3} \text{ dx }\]
\[ = \text{ ln} \left| x - 1 \right| + 2 \left[ \frac{\left( x - 1 \right)^{- 2 + 1}}{- 2 + 1} \right] + \left[ \frac{\left( x - 1 \right)^{- 3 + 1}}{- 3 + 1} \right] + C\]
\[ = \text{ ln} \left| x - 1 \right| - \frac{2}{\left( x - 1 \right)} - \frac{\left( x - 1 \right)^{- 2}}{2} + C\]
\[ = \text{ ln }\left| x - 1 \right| - \frac{2}{x - 1} - \frac{1}{2 \left( x - 1 \right)^2} + C\]
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