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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
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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