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∫ √ 16 X 2 + 25 D X

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प्रश्न

\[\int\sqrt{16 x^2 + 25} \text{ dx}\]
योग
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उत्तर

\[\int \sqrt{16 x^2 + 25} \text{ dx}\]
\[ = \int \sqrt{16\left( x^2 + \frac{25}{16} \right)}\text{ dx}\]
\[ = 4\int \sqrt{x^2 + \left( \frac{5}{4} \right)^2} \text{ dx}\]
\[ = 4\left[ \frac{x}{2}\sqrt{x^2 + \left( \frac{5}{4} \right)^2} + \frac{\left( \frac{5}{4} \right)^2}{2}\text{ ln }\left| x + \sqrt{x^2 + \left( \frac{5}{4} \right)^2} \right| \right] + C \left[ \because \int\sqrt{x^2 + a^2} \text{ dx} = \frac{1}{2}x\sqrt{x^2 + a^2} + \frac{1}{2} a^2 \text{ ln}\left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = 2x \sqrt{x^2 + \frac{25}{16}} + \frac{25}{8}\text{ ln }\left| x + \sqrt{x^2 + \frac{25}{16}} \right| + C\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 18 Indefinite Integrals
Exercise 19.28 | Q 8 | पृष्ठ १५४

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