Evaluvate the following intregals: 1 x(x-2)(x-4) dx - Vidyadip

Question

Evaluvate the following intregals:
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$

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Answer

Let $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-4}$ $\Rightarrow1=\text{A}(\text{x}-2)(\text{x}-4)+\text{B}(\text{x})(\text{x}-4)+\text{Cx}(\text{x}-2)$Put x = 0
$\Rightarrow1=8\text{A}\Rightarrow\text{A}=\frac{1}{8}$ Put x = 2 $\Rightarrow1=-4\text{B}\Rightarrow\text{B}=-\frac{1}{4}$ Put x = 4 $\Rightarrow1=8\text{C}\Rightarrow\text{C}=\frac{1}{8}$ So, $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{1}{8}\int\frac{\text{dx}}{\text{x}}+\Big(-\frac{1}{4}\Big)\int\frac{\text{dx}}{\text{x}-2}+\frac{1}{8}\int\frac{\text{dx}}{\text{x}-4}$ $=\frac{1}{8}\log|\text{x}|-\frac{1}{4}\log|\text{x}-2|+\frac{1}{8}\log|\text{x}-4|+\text{C}$ $=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$ $\text{I}=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$

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