Question

If $\int \frac{f\left(x\right)}{1-x^3}dx=\log \left|\frac{x^2+x+1}{x-1}\right|\frac{A}{948\sqrt{3}}ta{n}^{-1}\frac{2x+1}{\sqrt{3}}+C$ then A = ___,

where $f\left(x\right)$ is a polynomial of second degree in x such that $f\left(0\right)=f\left(1\right)=3f\left(2\right)=3$.

Answer 1

Let $f\left(x\right)=a{x}^2+bx+c$
From $f\left(0\right)=f\left(1\right)=3f\left(2\right)=3$, we get
$f\left(x\right)=-x^2+x+3$
Then
$I=\int \frac{-x^2+x+3}{1-x^3}=\int (\frac{2x+2}{x^2+x+1}-\frac{1}{x-1})dx=\int (\frac{2x+1}{x^2+x+1}+\frac{1}{x^2+x+1}-\frac{1}{x-1})dx=\log (x^2+x+1)+\frac{2{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)}{\sqrt{3}}-\log (x-1)+c=\log \frac{(x^2+x+1)}{x-1}+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c$
Therefore, $A=1896$