Question

Please answer my question
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Answer 1

Step-by-step explanation:

We have,

$\int \limits^{e}_{1} \bigg( \sqrt{x} + \frac{1}{ \sqrt{x} } \bigg)^{2}dx$

$= \int \limits^{e}_{1} x \: dx+\int \limits^{e}_{1} \frac{1}{ x } \: dx + 2 \int \limits^{e}_{1} \: dx$

$= \bigg [ \frac{ {x}^{2} }{2} \bigg ] ^{e}_{1} \:+ \bigg [ \ln(x )\bigg ] ^{e}_{1} + 2 \bigg [x \bigg ] ^{e}_{1}$

$= \frac{ {e}^{2} }{2} - \frac{1}{2} + \ln(e) - \ln(1) + 2 (e - 1)$

$= \frac{ {e}^{2} }{2} - \frac{1}{2} + 1 + 2 e - 2$

$= \frac{ {e}^{2} }{2} + 2e - \frac{3}{2}$

Answer 2

Answer:

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