∫ ( 2 X − 5 ) √ X 2 − 4 X + 3 D X - Mathematics

Question

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

Sum

Solution

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

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Indefinite Integral Problems

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Q 9 Q 8 Q 10

Chapter 19: Indefinite Integrals - Exercise 19.29 [Page 159]

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RD Sharma Mathematics [English] Class 12

Chapter 19 Indefinite Integrals
Exercise 19.29 | Q 9 | Page 159

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