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Question
Solution
\[\text{ Let I }= \int\frac{\left( 5x + 3 \right) dx}{\sqrt{x^2 + 4x + 10}}\]
\[\text{ Consider, }\]
\[5x + 3 = A \frac{d}{dx} \left( x^2 + 4x + 10 \right) + B\]
\[ \Rightarrow 5x + 3 = A \left( 2x + 4 \right) + B\]
\[ \Rightarrow 5x + 3 = \left( 2A \right) x + 4A + B\]
\[\text{Equating Coefficients of like terms}\]
\[\text{ 2 A} = 5\]
\[ \Rightarrow A = \frac{5}{2}\]
\[\text{ And }\]
\[4A + B = 3\]
\[ \Rightarrow 4 \times \frac{5}{2} + B = 3\]
\[ \Rightarrow B = - 7\]
\[ \therefore I = \frac{5}{2}\int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 10}} - 7\int\frac{dx}{\sqrt{x^2 + 4x + 10}}\]
\[ = \frac{5}{2}\int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 10}} - 7\int\frac{dx}{\sqrt{x^2 + 4x + 4 - 4 + 10}}\]
\[ = \frac{5}{2}\int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 10}} - 7\int\frac{dx}{\sqrt{\left( x + 2 \right)^2 + \left( \sqrt{6} \right)^2}}\]
\[\text{ Putting,} x^2 + 4x + 10 = t\]
\[ \Rightarrow \left( 2x + 4 \right) dx = dt\]
\[\text{ Then,} \]
\[I = \frac{5}{2}\int\frac{dt}{\sqrt{t}} - 7 \text{ log }\left| x + 2 + \sqrt{\left( x + 2 \right)^2 + 6} \right| + C\]
\[ = \frac{5}{2}\int t^{- \frac{1}{2}} dt - 7 \text{ log} \left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C\]
\[ = \frac{5}{2} \times 2\sqrt{t} - 7 \text{ log} \left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C\]
\[ = 5\sqrt{x^2 + 4x + 10} - 7 \text{ log }\left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C\]
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