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प्रश्न
उत्तर
\[\text{ Let I }= \int {1_{II} \cdot}\sqrt{a^2 {_I} + x^2} dx\]
\[ = \sqrt{a^2 + x^2} \int1 \text{ dx }- \int\left( \frac{d}{dx}\left( \sqrt{a^2 + x^2} \right) \int1 \text{ dx }\right)\text{ dx }\]
\[ = \sqrt{a^2 + x^2} \cdot x - \int\frac{1 \times 2x}{2 \sqrt{a^2 + x^2}} \cdot x \text{ dx }\]
\[ = \sqrt{a^2 + x^2} \cdot x - \int\left( \frac{x^2 + a^2 - a^2}{\sqrt{a^2 + x^2}} \right)\text{ dx }\]
\[ = x\sqrt{a^2 + x^2} - \int\sqrt{a^2 + x^2} dx + a^2 \int\frac{1}{\sqrt{a^2 + x^2}}\text{ dx }\]
\[ = x\sqrt{a^2 + x^2} - I + a^2 \int\frac{1}{\sqrt{a^2 + x^2}}dx\]
\[ \therefore 2I = x\sqrt{a^2 + x^2} + a^2 \text{ ln} \left| x + \sqrt{x^2 + a^2} \right|\]
\[ \Rightarrow I = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \text{ ln} \left| x + \sqrt{x^2 + a^2} \right| + C\]
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