If $I=\int \frac{x^3-1}{x^5+x^4+x+1}dx$ $=\frac{1}{4}\ln \left(f\left(x\right)\right)-\ln \left(g\left(x\right)\right)+c$ (where, $c$ is the constant of integration) and $f\left(0\right)=g\left(0\right)=1$, then the value of $f\left(1\right)\cdot g\left(1\right)$ is equal to

The given integral is $I=\int \frac{x^3-1}{\left(x+1\right)\left(x^4+1\right)}dx$
$=\int \frac{x^3+x^4-x^4-1}{\left(x+1\right)\left(x^4+1\right)}dx$
$=\int \frac{x^3\left(x+1\right)-\left(x^4+1\right)}{\left(x+1\right)\left(x^4+1\right)}dx$
$=\int \frac{x^3}{x^4+1}dx-\int \frac{dx}{x+1}$
$=\frac{1}{4}\int \frac{4x^{3}dx}{x^4+1}-\int \frac{dx}{x+1}$
$=\frac{1}{4}\ln \left(x^4+1\right)-\ln \left(1+x\right)+c$

Hence, $f\left(x\right)=x^4+1, g\left(x\right)=x+1$
$\therefore f\left(1\right)·g\left(1\right)=2\times 2=4$