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

Question

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

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Answer

Let $\text{I}=\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}$
we express
$\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}=\frac{\text{x}^2}{\text{x}^4+2\text{x}^2-\text{x}^2-2}$
$=\frac{\text{x}^2}{(\text{x}^2+2)(\text{x}^2-1)}$
$=\frac{\text{A}}{\text{x}^2+2}+\frac{\text{B}}{\text{x}^2-1}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2-1)+\text{B}(\text{x}^2+2)$
Equating the coefficient of x and constants, we get
1 = A + B and 0 = -A + 2B or
$\text{A}=\frac{2}{3}$ and $\text{B}=\frac{1}{3}$
$\therefore\text{I}=\int\Big(\frac{\frac{2}{3}}{\text{x}^2+2}+\frac{\frac{1}{3}}{\text{x}^2-1}\Big)\text{dx}$
$=\frac{2}{3}\int\frac{1}{\text{x}^2+2}\ \text{dx}+\frac{1}{3}\int\frac{1}{\text{x}^2-1}\ \text{dx}$
$=\frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\frac{1}{6}\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}=\frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\frac{1}{6}\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$

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