If f^{prime}(x)=x+frac{1}{x},then f(x)= | vedclass

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(B) Given that $f^{\prime}(x)=x+\frac{1}{x}$.To find $f(x)$,we integrate both sides with respect to $x$:$f(x) = \int f^{\prime}(x) dx = \int \left(x +

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

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(B) Given that $f^{\prime}(x)=x+\frac{1}{x}$.To find $f(x)$,we integrate both sides with respect to $x$:$f(x) = \int f^{\prime}(x) dx = \int \left(x + \frac{1}{x}\right) dx$.Using the linearity property of integration:$f(x) = \int x dx + \int \frac{1}{x} dx$.Applying the standard integration formulas $\int x^n dx = \frac{x^{n+1}}{n+1}$ and $\int \frac{1}{x} dx = \log |x| + C$:$f(x) = \frac{x^2}{2} + \log |x| + C$.

If $f^{\prime}(x)=x+\frac{1}{x}$,then $f(x)=$

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