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Question
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Solution
\[\text{ Let I }= \int\frac{\left( 2x + 5 \right) dx}{\sqrt{x^2 + 2x + 5}}\]
\[\text{ Consider,} \]
\[2x + 5 = A \frac{d}{dx} \left( x^2 + 2x + 5 \right) + B\]
\[ \Rightarrow 2x + 5 = A \left( 2x + 2 \right) + B\]
\[ \Rightarrow 2x + 5 = \left( 2A \right) x + 2A + B\]
\[\text{Equating Coefficients of like terms}\]
\[2A = 2 \Rightarrow A = 1\]
\[\text{ And }\]
\[ 2A + B = 5 \Rightarrow B = 3\]
\[ \therefore I = \int\left( \frac{2x + 2 + 3}{\sqrt{x^2 + 2x + 5}} \right) dx\]
\[ = \int\frac{\left( 2x + 2 \right) dx}{\sqrt{x^2 + 2x + 5}} + 3\int\frac{dx}{\sqrt{x^2 + 2x + 5}}\]
\[\text{ let x}^2 + 2x + 5 = t\]
\[ \Rightarrow \left( 2x + 2 \right) dx = dt\]
\[\text{ Then,} \]
\[I = \int\frac{dt}{\sqrt{t}} + 3\int\frac{dx}{\sqrt{x^2 + 2x + 1 + 4}}\]
\[ = \int t^{- \frac{1}{2}} dt + 3 \int\frac{dx}{\sqrt{\left( x + 1 \right)^2 + 2^2}}\]
\[ = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + 3 \text{ log }\left| x + 1 + \sqrt{\left( x + 1 \right)^2 + 4} \right| + C\]
\[ = 2\sqrt{t} + 3 \text{ log }\left| x + 1 + \sqrt{x^2 + 2x + 5} \right| + C\]
\[ = 2\sqrt{x^2 + 2x + 5} + 3 \text{ log }\left| x + 1 + \sqrt{x^2 + 2x + 5} \right| + C\]
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