If f^{prime}(x)=x-frac{5}{x^5} and f(1)=4,the | vedclass

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(C) Given $f^{\prime}(x)=x-\frac{5}{x^5}$.Integrating both sides with respect to $x$,we get:$f(x) = \int \left(x - 5x^{-5}\right) dx$$f(x) = \frac{x^2

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$\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}$

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(C) Given $f^{\prime}(x)=x-\frac{5}{x^5}$.Integrating both sides with respect to $x$,we get:$f(x) = \int \left(x - 5x^{-5}\right) dx$$f(x) = \frac{x^2}{2} - 5 \left( \frac{x^{-4}}{-4} \right) + c$$f(x) = \frac{x^2}{2} + \frac{5}{4x^4} + c$Given $f(1) = 4$,substitute $x=1$:$4 = \frac{1^2}{2} + \frac{5}{4(1)^4} + c$$4 = \frac{1}{2} + \frac{5}{4} + c$$4 = \frac{2+5}{4} + c$$4 = \frac{7}{4} + c$$c = 4 - \frac{7}{4} = \frac{16-7}{4} = \frac{9}{4}$Thus,$f(x) = \frac{x^2}{2} + \frac{5}{4x^4} + \frac{9}{4}$.

If $f^{\prime}(x)=x-\frac{5}{x^5}$ and $f(1)=4$,then $f(x)$ is

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