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प्रश्न
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
योग
उत्तर
\[I = \int\sqrt{x^2 - 2x}dx\]
\[\Rightarrow I = \int\sqrt{x^2 - 2x + 1 - 1}\text{ dx}\]
\[ \Rightarrow I = \int\sqrt{(x - 1 )^2 - 1^2}dx\]
\[ \because \int\sqrt{x^2 - a^2}dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\text{ ln}\left( \left| x + \sqrt{x^2 - a^2} \right| \right) + c\]
\[ \therefore I = \frac{(x - 1)}{2}\sqrt{(x - 1 )^2 - 1} - \frac{1}{2}\text{ ln}\left| \left( x - 1 \right) + \sqrt{x^2 - 2x} \right| + c\]
\[ \Rightarrow I = \int\sqrt{(x - 1 )^2 - 1^2}dx\]
\[ \because \int\sqrt{x^2 - a^2}dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\text{ ln}\left( \left| x + \sqrt{x^2 - a^2} \right| \right) + c\]
\[ \therefore I = \frac{(x - 1)}{2}\sqrt{(x - 1 )^2 - 1} - \frac{1}{2}\text{ ln}\left| \left( x - 1 \right) + \sqrt{x^2 - 2x} \right| + c\]
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