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प्रश्न
उत्तर
\[Let\ x = a \tan\ t . Then, dx = a\ \sec^2 t\ dt\]
\[When\ x = 0, t = 0\ and\ x = a, t = \frac{\pi}{4}\]
\[ \therefore I = \int_0^a \frac{x}{\sqrt{a^2 + x^2}} d\ x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{4} \frac{a \tan t}{\sqrt{a^2 + a^2 \tan^2 t}}a \sec^2 t\ d t\]
\[ = \int_0^\frac{\pi}{4} \frac{\left( a \tan t \right) a \sec^2 t}{a \sec t} dt\]
\[ = \int_0^\frac{\pi}{4} a \tan t \sec t\ dt\]
\[ = a \left[ \sec t \right]_0^\frac{\pi}{4} \]
\[ = a\left( \sqrt{2} - 1 \right)\]
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