मराठी

∫ √ X 2 + X + 1 D X

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प्रश्न

\[\int\sqrt{x^2 + x + 1} \text{ dx}\]
बेरीज
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उत्तर

\[\int \sqrt{x^2 + x + 1} \text{ dx}\]
\[ = \int \sqrt{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1} \text{ dx}\]
\[ = \int \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \left( \frac{x + \frac{1}{2}}{2} \right) \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{3}{8}\text{ log } \left| \left( x + \frac{1}{2} \right) + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C \left[ \because \int\sqrt{x^2 + a^2}dx = \frac{1}{2}x\sqrt{x^2 + a^2} + \frac{1}{2} a^2 \text{ ln }\left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = \left( \frac{2x + 1}{4} \right) \sqrt{x^2 + x + 1} + \frac{3}{8}\text{ log }\left| \left( 2x + 1 \right) + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C\]

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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 2 | पृष्ठ १५४

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