Let f(x)=intleft(frac{2 x^3-3 x^2+4 x-5}{x^2} | vedclass

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(C) Given: $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$We simplify the integrand: $f(x)=\int\left(2 x-3+\frac{4}{x}-\frac{5}{x^2}\right)

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$9+4 \log 5$

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(C) Given: $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$We simplify the integrand: $f(x)=\int\left(2 x-3+\frac{4}{x}-\frac{5}{x^2}\right) d x$Integrating term by term: $f(x)=x^2-3 x+4 \log |x|+\frac{5}{x}+C$Given $f(1)=1$,we substitute $x=1$: $1^2-3(1)+4 \log(1)+\frac{5}{1}+C=1$$1-3+0+5+C=1 \Rightarrow 3+C=1 \Rightarrow C=-2$Thus,$f(x)=x^2-3 x+4 \log |x|+\frac{5}{x}-2$Now,calculate $f(5)$: $f(5)=5^2-3(5)+4 \log 5+\frac{5}{5}-2$$f(5)=25-15+4 \log 5+1-2$$f(5)=9+4 \log 5$

Let $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$ and $f(1)=1$. Then $f(5)=$

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