Question

If $\int \frac{x^2-1}{x^4+x^2+1}dx=\frac{1}{2}\ln \left|\frac{x^2-x+K}{x^2+x+K}\right|+C$, Then value of $K$ is (where $K$ is fixed constant and $C$ is integration constant)

• A. 1
• B. 1.00
• C. 1.0

Answer 1

Answer: (A), (B), (C)

$I=\int \frac{x^2-1}{x^4+x^2+1}dx$
Dividing $N^r$ and $D^r$ by $x^2$
$=\int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}dx$
$=\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-1^2}dx$
Let $x+\frac{1}{x}=u$
$\Rightarrow (1-\frac{1}{x^2})dx=du$
$\Rightarrow I=\int \frac{du}{u^2-1^2}=\frac{1}{2\left(1\right)}\ln \left|\frac{u-1}{u+1}\right|+C$
$=\frac{1}{2}\ln \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+C$
$\Rightarrow$ $I=\frac{1}{2}\ln \left|\frac{x^2-x+1}{x^2+x+1}\right|+C$
$\therefore K=1$