Question

Evaluate $\int \frac{f\left(x\right)}{x^3-1}dx$, where $f\left(x\right)$ is a polynomial of degree $2$ in $x$ such that $f\left(0\right)=f\left(1\right)=3f\left(2\right)=-3$

• A. $I=-\log |x-1|+\log |x^2+x+1|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$
• B. $I=\log |x-1|+\log |x^2+x+1|+\frac{2}{\sqrt{3}}{\cot }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$
• C. $I=-\log |x-1|+\log |x^2+x+1|+\frac{}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+1}{\sqrt{3}}\right)+C$
• D. $I=-\log |x-1|-\log |x^2+x+1|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$

Answer 1

The correct option is A $I=-\log |x-1|+\log |x^2+x+1|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$

Let $f\left(x\right)=a{x}^2+bx+c$
Given, $f\left(0\right)=f\left(1\right)=3f\left(2\right)=-3$
$\therefore f\left(0\right)=c=-3$
$f\left(1\right)=a+b+c=-3$
$3f\left(2\right)=3(4a+2b+c)=-3$
On solving, we get $a=1$, $b=-1$, $c=-3$
$\therefore f\left(x\right)=x^2-x-3$
$⟹I=\int \frac{f\left(x\right)}{x^3-1}dx=\int \frac{x^2-x-3}{(x-1)(x^2+x+1)}dx$
Using partial fractions, we get
$\frac{(x^2-x-3)}{(x-1)(x^2+x+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+x+1)}$
$\Rightarrow x^2-x-3=(A+B)x^2+(A-B+C)x+(A-C)$
$\Rightarrow A+B=1$
$A-B+C=-1$
$A-C=-3$
On solving, we get $A=-1$, $B=2$, $C=2$
$\therefore I=\int -\frac{1}{x-1}dx+\int \frac{(2x+2)}{(x^2+x+1)}dx$
$=-\log |x-1|+\int \frac{(2x+1)dx}{x^2+x+1}+\int \frac{1dx}{x^2+x+1}$
Put $x^2+x+1=t$
$\Rightarrow (2x+1)dx=dt$
$=-\log |x-1|+\int \frac{dt}{t}+\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=-\log |x-1|+\log |x^2+x+1|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)$
$=-\log |x-1|+\log |x^2+x+1|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$