Evaluate the following intregals: x^2+6x-8 x^3-4x dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}^2+6\text{x}-8}{\text{x}^3-4\text{x}}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}^2+6\text{x}-8}{\text{x}^3-4\text{x}}\ \text{dx}$ $\Rightarrow\text{I}=\int\frac{\text{x}^2+6\text{x}-8}{\text{x}(\text{x}+2)(\text{x}-2)}\text{ dx}$Now,
Let $\frac{\text{x}^2+6\text{x}-8}{\text{x}(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+2}+\frac{\text{C}}{\text{x}-2}$
$\Rightarrow\text{x}^2+6\text{x}-8=\text{A}(\text{x}^2-4)+\text{B}(\text{x}-2)\text{x}+\text{C}(\text{x}+2)\text{x}$ Put x = 0 ⇒ -8 = -4A ⇒ A = 2 Put x = -2 ⇒ -16 = 8B ⇒ B = -2 Put x = 2 ⇒ 8 = 8C ⇒ C = 1 Thus,$\text{I}=\int\frac{2\text{dx}}{\text{x}}-\int\frac{2\text{dx}}{\text{x}+2}+\int\frac{\text{dx}}{\text{x}-2}$
$=2\log|\text{x}|-2\log|\text{x}+2|+\log|\text{x}-2|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{x}^2(\text{x}-2)}{(\text{x}+2)^2}\Big|+\text{C}$

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