Question

\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{  dx }\]

Answer 1

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