Question

$\int \frac{x^2-1}{x^4+x^2+1}dx$ is equal to

• A. $\log (x^4+x^2+1)+c$
• B. $\log \frac{x^2-x+1}{x^2+x+1}+c$
• C. $\frac{1}{2}\log \frac{x^2-x+1}{x^2+x+1}+c$
• D. $\frac{1}{2}\log \frac{x^2+x+1}{x^2-x+1}+c$

Answer 1

The correct option is C $\frac{1}{2}\log \frac{x^2-x+1}{x^2+x+1}+c$

Let $I=\int \frac{x^2-1}{x^4+x^2+1}dx$
$\Rightarrow I=\int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}+1-1}dx$
$=\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-1}dx$
Put $x+\frac{1}{x}=t\Rightarrow 1-\frac{1}{x^2}dx=dt$
$\therefore I=\int \frac{1}{t^2-1}dt$
$=\frac{1}{2\times 1}\log \frac{t-1}{t+1}+c$
$=\frac{1}{2}\log \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}+c$
$=\frac{1}{2}\log \frac{x^2-x+1}{x^2+x+1}+c$