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Question
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Solution
\[\text{ Let I } = \int\frac{\left( x + 1 \right)}{x^2 + 4x + 5}dx\]
\[\text{ and let} \left( x + 1 \right) = A\frac{d}{dx}\left( x^2 + 4x + 5 \right) + B\]
\[ \Rightarrow x + 1 = A \left( 2x + 4 \right) + B\]
\[ \Rightarrow x + 1 = \left( 2A \right)x + 4A + B\]
\[\text{Equating the coefficients of like terms}\]
\[2A = 1\]
\[ \Rightarrow A = \frac{1}{2}\]
\[\text{ and }\ 4A + B = 1\]
\[ \Rightarrow 4 \times \frac{1}{2} + B = 1\]
\[ \Rightarrow B = - 1\]
\[ \therefore \left( x + 1 \right) = \frac{1}{2} \left( 2x + 4 \right) - 1\]
\[ \therefore I = \int\left[ \frac{\frac{1}{2}\left( 2x + 4 \right) - 1}{x^2 + 4x + 5} \right]dx\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 4 \right)}{x^2 + 4x + 5}dx - \int\frac{1}{x^2 + 4x + 5}dx\]
\[\text{ Putting x}^2 + 4x + 5 = t\]
\[ \Rightarrow \left( 2x + 4 \right) dx = dt\]
\[ \therefore I = \frac{1}{2}\int\frac{1}{t}dt - \int\frac{1}{x^2 + 4x + 4 + 1}dx\]
\[ = \frac{1}{2}\int\frac{dt}{t} - \int\frac{1}{\left( x + 2 \right)^2 + 1^2}dx \]
\[ = \frac{1}{2} \text{ ln } \left| t \right| - \tan^{- 1} \left( \frac{x + 2}{1} \right) + C............. \left[ \because \int\frac{1}{x^2 + a^2}dx = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right]\]
\[ = \frac{1}{2} \text{ ln }\left| x^2 + 4x + 5 \right| - \tan^{- 1} \left( x + 2 \right) + C ...................\left[ \because t = x^2 + 4x + 5 \right]\]
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