Question

\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{  dx }\]

 

Answer 1

\[\text{ Let I }= \int \left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{  dx }\]
\[ = \int \left( 2x - 4 - 1 \right) \sqrt{x^2 - 4x + 3} \text{  dx }\]
\[ = \int\left( 2x - 4 \right) \sqrt{x^2 - 4x + 3} \text{  dx }- \int\sqrt{x^2 - 4x + 3} \text{  dx }\]
\[ = \int\left( 2x - 4 \right) \sqrt{x^2 - 4x + 3} \text{  dx }- \int \sqrt{x^2 - 4x + 4 - 4 + 3} \text{  dx }\]
\[ = \int\left( 2x - 4 \right) \sqrt{x^2 - 4x + 3} \text{  dx }- \int \sqrt{\left( x - 2 \right)^2 - 1^2} \text{  dx }\]
\[\text{ Let x}^2 - 4x + 3 = t\]
\[ \Rightarrow \left( 2x - 4 \right)dx = dt\]
\[ \therefore I = \int\sqrt{t}\text{  dt }- \int\sqrt{\left( x - 2 \right)^2 - 1^2} dx\]
\[ = \frac{2}{3} t^\frac{3}{2} - \left[ \frac{x - 2}{2} \sqrt{\left( x - 2 \right)^2 - 1^2} - \frac{1^2}{2}\text{ log }\left| \left( x - 2 \right) + \sqrt{\left( x - 2 \right)^2 - 1} \right| \right] + C\]
\[ = \frac{2}{3} \left( x^2 - 4x + 3 \right)^\frac{3}{2} - \left( \frac{x - 2}{2} \right) \sqrt{x^2 - 4x + 3} + \frac{1}{2}\text{ log }\left| \left( x - 2 \right) + \sqrt{x^2 - 4x + 3} \right| + C\]