int_1^4 left(x + sqrt{x} + frac{1}{x}right) d | vedclass

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(A) Let $I = \int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx$. First, evaluate the first integral: $\int_1^4 \left(x + x^{1

Vedclass · Accepted Answer

$\frac{79}{6}$

Vedclass · Answer

(A) Let $I = \int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx$. First, evaluate the first integral: $\int_1^4 \left(x + x^{1/2} + \frac{1}{x}\right) dx = \left[ \frac{x^2}{2} + \frac{2}{3}x^{3/2} + \log |x| \right]_1^4$. $= \left( \frac{16}{2} + \frac{2}{3}(8) + \log 4 \right) - \left( \frac{1}{2} + \frac{2}{3} + 0 \right) = \left( 8 + \frac{16}{3} + 2 \log 2 \right) - \left( \frac{7}{6} \right) = \frac{40}{3} + 2 \log 2 - \frac{7}{6} = \frac{80 - 7}{6} + 2 \log 2 = \frac{73}{6} + 2 \log 2$. Now, evaluate the second integral: $\int_1^{2 \log 2} dx = [x]_1^{2 \log 2} = 2 \log 2 - 1$. Subtracting the two results: $I = \left( \frac{73}{6} + 2 \log 2 \right) - (2 \log 2 - 1) = \frac{73}{6} + 1 = \frac{73 + 6}{6} = \frac{79}{6}$.

$\int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx = $

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