If f'(x) = frac{1}{x} + x and f(1) = frac{5}{ | vedclass

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(A) Given $f'(x) = \frac{1}{x} + x$.To find $f(x)$,we integrate $f'(x)$ with respect to $x$:$f(x) = \int (\frac{1}{x} + x) dx = \log|x| + \frac{x^2}{2

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$\log x + \frac{x^2}{2} + 2$

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(A) Given $f'(x) = \frac{1}{x} + x$.To find $f(x)$,we integrate $f'(x)$ with respect to $x$:$f(x) = \int (\frac{1}{x} + x) dx = \log|x| + \frac{x^2}{2} + C$.We are given $f(1) = \frac{5}{2}$. Substituting $x = 1$ into the equation:$f(1) = \log(1) + \frac{1^2}{2} + C = \frac{5}{2}$.Since $\log(1) = 0$,we have $0 + \frac{1}{2} + C = \frac{5}{2}$.$C = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2$.Substituting the value of $C$ back into the expression for $f(x)$,we get:$f(x) = \log x + \frac{x^2}{2} + 2$.

If $f'(x) = \frac{1}{x} + x$ and $f(1) = \frac{5}{2}$,then $f(x) = $

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