Evaluate the following integrals: x^2-1 x^4+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^2-1}{\text{x}^4+1}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}^2-1}{\text{x}^4+1}\ \text{dx}$
Dividing numerator and denominator bt $x^2$
$\therefore\text{I}=\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)}{\text{x}^2+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2}$
Let $\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2-2}$
$=\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{t}-\sqrt{2}}{\text{t}+\sqrt{2}}\Big|+\text{C}$
So,
$\text{I}=\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{x}^2+1-\sqrt{2}\text{x}}{\text{x}^2+1+\sqrt{2}\text{x}}\Big|+\text{C}$

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