Question
Evaluate the following intregals:
$\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$
$\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$We express
$\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}=\frac{\text{x}^2}{\text{x}^4-4\text{x}^2+3\text{x}^2-12}$
$=\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}$
$=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2+3)+\text{B}(\text{x}^2-4)$
Equating the coefficients of $x^2$ and constant, we get
1 = A + B and 0 = 3A - 4B or
$\text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\therefore\text{I}=\int\bigg(\frac{\frac{4}{7}}{\text{x}^2-4}+\frac{\frac{3}{7}}{\text{x}^2+3}\bigg)\text{dx}$
$=\frac{4}{7}\int\frac{1}{\text{x}^2-4}\ \text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+3}\text{dx}$
$=\frac{4}{7}\times\frac{1}{4}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}=\frac{\text{x}^2}{\text{x}^4-4\text{x}^2+3\text{x}^2-12}$
$=\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}$
$=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2+3)+\text{B}(\text{x}^2-4)$
Equating the coefficients of $x^2$ and constant, we get
1 = A + B and 0 = 3A - 4B or
$\text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\therefore\text{I}=\int\bigg(\frac{\frac{4}{7}}{\text{x}^2-4}+\frac{\frac{3}{7}}{\text{x}^2+3}\bigg)\text{dx}$
$=\frac{4}{7}\int\frac{1}{\text{x}^2-4}\ \text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+3}\text{dx}$
$=\frac{4}{7}\times\frac{1}{4}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
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