Question

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

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

Answer 1

Answer: (D)

$\log |x^2+x+1|-\log |x-1|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c$

$f\left(x\right)$ is degree 2

Let $f\left(x\right)=a{x}^2+bx+c$

$c=-3;a=1;b=-1$

$\int \frac{x^2-x-3}{(x^3-1)}dx=\int \frac{x^2-x-3}{(x-1)(x^2+x+1)}$

$\frac{x^2-x-3}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$

$x^2-x-3=A(x^2+x+1)+(Bx+c)(x-1)]A=-1;B=2;C=2$

$\int \frac{x^2-x-3}{(x-1)(x^2+x+1)}dx=\int \frac{-1}{x-1}dx+2\int \frac{(x+1)}{x^2+x+1}dx$

$=-log\mid x-1\mid +2\int \frac{x+1}{(x^2+x+1)}dx$

$-log\mid x-1\mid +2[\int \frac{\frac{1}{2}(2x+1)}{(x^2+x+1)}dx+\int \frac{1}{2}\frac{1}{x^2+x+1}dx]$

$=-log\mid x-1\mid +\int \frac{(2x+1)}{(x^2+x+1)}dx+\int \frac{1}{x^2+x+1}\cdot dx.$

$=-log\mid x-1\mid +log\mid x^2+x+1\mid +\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c.$