Given:

$f\left(x\right)=\int \left(\frac{2x^3-3x^2+4x-5}{x^2}\right)dx$

$\Rightarrow f\left(x\right)=\int \left(2x-3+\frac{4}{x}-\frac{5}{x^2}\right)dx$

$\Rightarrow f\left(x\right)=x^2-3x+4\log x+\frac{5}{x}+C$

where, $C$ is the constant of integartion.

$f\left(1\right)=1$

$\Rightarrow 1-3+4\times 0+5+C=1$

$\Rightarrow C=-2$

So,

$f\left(x\right)=x^2-3x+4\log x+\frac{5}{x}-2$

$\Rightarrow f\left(5\right)=25-15+4\log 5+1-2$

$\Rightarrow f\left(5\right)=9+4\log 5$