Question

\[\int\frac{2x + 1}{\sqrt{x^2 + 4x + 3}} \text{ dx }\]

Answer 1

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