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Question
\[\int\frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} dx\]
Sum
Solution
\[\text{Let I }= \int \left[ \frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} \right]dx\]
Putting x + 1 = t
⇒ x = t – 1
& dx = dt
\[\therefore I = \int\left[ \frac{\left( t - 1 \right)^2 + 3 \left( t - 1 \right) - 1}{t^2} \right]dt\]
\[ = \int \left( \frac{t^2 - 2t + 1 + 3t - 3 - 1}{t^2} \right)dt\]
\[ = \int\left( \frac{t^2 + t - 3}{t^2} \right)dt\]
\[ = \int\left( 1 + \frac{1}{t} - 3 t^{- 2} \right)dt\]
\[ = t + \text{log} \left| t \right| - 3\left( \frac{t^{- 2 + 1}}{- 2 + 1} \right) + C\]
\[ = t + \text{log}\left| t \right| + \frac{3}{t} + C\]
\[ = x + 1 + \text{log} \left| x + 1 \right| + \frac{3}{x + 1} + C \left[ \because t = x + 1 \right]\]
Let C + 1 = C′
\[= x + \text{log} \left( x + 1 \right) + \frac{3}{x + 1} + C\prime\]
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