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Question
Solution
\[\int\left[ \frac{x + 3}{\left( x + 1 \right)^4} \right]dx\]
\[ = \int\left[ \frac{x + 1 + 2}{\left( x + 1 \right)^4} \right]dx\]
\[ = \int\left[ \frac{\left( x + 1 \right)}{\left( x + 1 \right)^4} + \frac{2}{\left( x + 1 \right)^4} \right]dx\]
\[ = \int\frac{dx}{\left( x + 1 \right)^3} + 2\int\frac{dx}{\left( x + 1 \right)^4}\]
\[ = \int \left( x + 1 \right)^{- 3} dx + 2\int \left( x + 1 \right)^{- 4} dx\]
\[ = \left[ \frac{\left( x + 1 \right)^{- 3 + 1}}{- 3 + 1} \right] + 2\left[ \frac{\left( x + 1 \right)^{- 4 + 1}}{- 4 + 1} \right] + C\]
\[ = - \frac{1}{2} \left( x + 1 \right)^{- 2} - \frac{2}{3} \left( x + 1 \right)^{- 3} + C\]
\[ = - \frac{1}{2 \left( x + 1 \right)^2} - \frac{2}{3 \left( x + 1 \right)^3} + C\]
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If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then