Question

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

Answer 1

The given integral is $I=\int \frac{x^{3} - 1}{\left(x + 1\right) \left(x^{4} + 1\right)}dx$
$= \int \frac{x^{3} + x^{4} - x^{4} - 1}{\left(x + 1\right) \left(x^{4} + 1\right)}dx$
$=\int \frac{x^{3} \left(x + 1\right) - \left(x^{4} + 1\right)}{\left(x + 1\right) \left(x^{4} + 1\right)}dx$
$=\int \frac{x^{3}}{x^{4} + 1}dx- \int \frac{d x}{x + 1}$
$=\frac{1}{4} \int \frac{4 x^{3} d x}{x^{4} + 1}- \int \frac{d x}{x + 1}$
$=\frac{1}{4}ln\left(x^{4} + 1\right)-ln\left(1 + x\right)+c$
Hence, $f\left(x\right)=x^{4}+1,g\left(x\right)=x+1$
$\therefore f\left(1\right)\cdot g\left(1\right)=2\times 2=4$