मराठी

∫ √ 4 X 2 − 5 D X

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प्रश्न

\[\int\sqrt{4 x^2 - 5}\text{ dx}\]
बेरीज
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उत्तर

\[\int \sqrt{4 x^2 - 5}\text{ dx}\]
\[ = \int \sqrt{4\left( x^2 - \frac{5}{4} \right)} \text{ dx}\]
\[ = 2\int \sqrt{x^2 - \left( \frac{\sqrt{5}}{2} \right)^2} \text{ dx}\]
\[ = 2\left[ \frac{x}{2}\sqrt{x^2 - \frac{5}{4}} - \frac{5}{8}\text{ ln }\left| x + \sqrt{x^2 - \frac{5}{4}} \right| \right] + C \left[ \because \int\sqrt{x^2 - a^2} \text{ dx}= \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{1}{2} a^2 \text{ ln}\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]
\[ = x \sqrt{x^2 - \frac{5}{4}} - \frac{5}{4}\text{ ln }\left| x + \sqrt{x^2 - \frac{5}{4}} \right| + C\]

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पाठ 18: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 18 Indefinite Integrals
Exercise 19.28 | Q 9 | पृष्ठ १५४

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