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Mathematics · JEE | INTEGRALS
हिन्दीEvaluate \( \int \frac{x}{\sqrt{x+4}} \, dx \) for \( x > 0 \).
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Correct answer · Option B
\( \frac{2}{3}(x-8)\sqrt{x+4} + C \)
Explanation
Put \( x+4 = t^2 \), so \( dx = 2t dt \) and \( x = t^2 - 4 \). The integral evaluates to \( \int \frac{t^2-4}{t} (2t dt) = 2\int (t^2-4) dt = 2(\frac{t^3}{3} - 4t) = \frac{2}{3}t(t^2-12) = \frac{2}{3}\sqrt{x+4}(x+4-12) = \frac{2}{3}(x-8)\sqrt{x+4} + C \).
