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Question
Evaluate:
`∫(sqrtx + 1/sqrtx)^2 dx`
Evaluate
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Solution
`∫(sqrtx + 1/sqrtx)^2 dx`
= `∫(x + 1/x + 2sqrtx 1/sqrtx)dx`
= `∫(x + 1/x + 2)dx` ...[∵ ∫`1/x` = logx + c ]
= `x^2/2 + log x + 2x + c`
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