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प्रश्न
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
योग
उत्तर
\[\int \frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}}dx\]
\[ = \int \left( \frac{1}{\sqrt{x}} + \frac{x}{\sqrt{x}} + 2\frac{\sqrt{x}}{\sqrt{x}} \right)dx\]
\[ = \int\left( x^{- \frac{1}{2}} + x^\frac{1}{2} + 2 \right)dx\]
\[ = \left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \left[ \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + 2x + C\]
\[ = 2\sqrt{x} + \frac{2}{3} x^\frac{3}{2} + 2x + C\]
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