$\text{ If }f\left(x\right)=\int \frac{5x^4+4x^5}{{\left(x^5+x+1\right)}^2}dx\text{ and }f\left(0\right)=0,\text{ then }f\left(1\right)=$

$f\left(x\right)=\int \frac{5x^4+4x^5}{{\left(x^5+x+1\right)}^2}dx=\int \frac{5x^2+4x^5}{x^{10}{\left(1+\frac{1}{x^4}+\frac{1}{x^5}\right)}^2}$

$=\int \frac{\frac{5}{x^6}+\frac{4}{x^5}}{{\left(1+\frac{1}{x^4}+\frac{1}{x^5}\right)}^2}dx\left[\begin{array}{l}Put1+\frac{1}{x^4}+\frac{1}{x^5}=t\\ \Rightarrow \left(\frac{-4}{x^5}-\frac{5}{x^6}\right)dx=ct\end{array}\right]$

$\int -\frac{1}{t^2}dt=\frac{1}{t}+C$

$f\left(x\right)=\frac{1}{\left(1+\frac{1}{x^4}+\frac{1}{x^5}\right)}+C=\frac{x^5}{\left(x^5+x+1\right)}+C$

$f\left(x\right)=\frac{x^5}{x^5+x+1}\left[f\left(0\right)=0+C\Rightarrow 0+C\Rightarrow C=0\right]$

$f\left(1\right)=\frac{1}{3}=0.33$