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Antiderivative

If ∫x2-1x4+x2...

Question

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

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Solution

$I = \int \frac{x^{2} - 1}{x^{4} + x^{2} + 1} d x$
Dividing $N^{r}$ and $D^{r}$ by $x^{2}$
$= \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} d x$
$= \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 1^{2}} d x$
Let $x + \frac{1}{x} = u$
$\Rightarrow (1 - \frac{1}{x^{2}}) d x = d u$
$\Rightarrow I = \int \frac{d u}{u^{2} - 1^{2}} = \frac{1}{2 (1)} ln ∣ ∣ ∣ \frac{u - 1}{u + 1} ∣ ∣ ∣ + C$
$= \frac{1}{2} ln ∣ ∣ ∣ ∣ ∣ \frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1} ∣ ∣ ∣ ∣ ∣ + C$
$\Rightarrow$ $I = \frac{1}{2} ln ∣ ∣ ∣ \frac{x^{2} - x + 1}{x^{2} + x + 1} ∣ ∣ ∣ + C$
$∴ K = 1$

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