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प्रश्न
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उत्तर
\[\text{ Let I }= \int \left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[ = \int \left( 2x - 4 - 1 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[ = \int\left( 2x - 4 \right) \sqrt{x^2 - 4x + 3} \text{ dx }- \int\sqrt{x^2 - 4x + 3} \text{ dx }\]
\[ = \int\left( 2x - 4 \right) \sqrt{x^2 - 4x + 3} \text{ dx }- \int \sqrt{x^2 - 4x + 4 - 4 + 3} \text{ dx }\]
\[ = \int\left( 2x - 4 \right) \sqrt{x^2 - 4x + 3} \text{ dx }- \int \sqrt{\left( x - 2 \right)^2 - 1^2} \text{ dx }\]
\[\text{ Let x}^2 - 4x + 3 = t\]
\[ \Rightarrow \left( 2x - 4 \right)dx = dt\]
\[ \therefore I = \int\sqrt{t}\text{ dt }- \int\sqrt{\left( x - 2 \right)^2 - 1^2} dx\]
\[ = \frac{2}{3} t^\frac{3}{2} - \left[ \frac{x - 2}{2} \sqrt{\left( x - 2 \right)^2 - 1^2} - \frac{1^2}{2}\text{ log }\left| \left( x - 2 \right) + \sqrt{\left( x - 2 \right)^2 - 1} \right| \right] + C\]
\[ = \frac{2}{3} \left( x^2 - 4x + 3 \right)^\frac{3}{2} - \left( \frac{x - 2}{2} \right) \sqrt{x^2 - 4x + 3} + \frac{1}{2}\text{ log }\left| \left( x - 2 \right) + \sqrt{x^2 - 4x + 3} \right| + C\]
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संबंधित प्रश्न
Evaluate the following integrals:
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
