Correct Answer - Option 4 :
\(\frac{1}{8}\log \left| x \right| - \frac{1}{4}\log \left| {x - 2} \right| + \frac{1}{8}\log \left| {x - 4} \right| + C\)
Concept:
Partial Fraction:
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Factors in the denominator
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Corresponding Partial Fraction
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(x - a)
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\(\frac{A}{{x - a}}\)
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(x - b)2
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\(\frac{A}{{x - b}} + \frac{B}{{{{\left( {x - b} \right)}^2}}}\)
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(x - a) (x - b)
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\(\frac{A}{{\left( {x - a} \right)}} + \frac{B}{{\left( {x - b} \right)}}\)
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(x - c)3
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\(\frac{A}{{x - c}} + \frac{B}{{{{\left( {x - c} \right)}^2}}} + \frac{C}{{{{\left( {x - c} \right)}^3}}}\)
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(x - a) (x2 - a)
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\(\frac{A}{{\left( {x - a} \right)}} + \frac{{Bx + C}}{{\left( {{x^2} - a} \right)}}\)
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(ax2 + bx + c)
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\(\frac{{Ax + B}}{{\left( {a{x^2} + bx + c} \right)}}\)
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Calculation:
Here we have to find the value of \(\smallint \frac{{dx}}{{x\left( {x - 2} \right)\left( {x - 4} \right)}}\)
Let \(\frac{1}{{x\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{A}{x} + \frac{B}{{x - 2}} + \frac{C}{{x - 4}}\)
⇒ 1 = A (x - 2) (x - 4) + B [x (x - 4)] + C [x (x - 2)] ----(1)
By putting x = 0 on both sides of (1) we get A = 1/8
By putting x = 2 on both sides of (1) we get B = - 1/4
By putting x = 4 on both sides of (1) we get C = 1/8
\(\Rightarrow \frac{1}{{x\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{1}{{8x}} - \frac{1}{{4\;\left( {x - 2} \right)}} + \frac{1}{{8\left( {x - 4} \right)}}\)
\(\Rightarrow \smallint \frac{{dx}}{{x\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{1}{8}\;\smallint \frac{{dx}}{x} - \frac{1}{4}\;\smallint \frac{{dx}}{{x - 2}} + \frac{1}{8}\;\smallint \frac{{dx}}{{x - 4}}\)
As we know that \(\smallint \frac{{dx}}{x} = \log \left| x \right|\; + C\) where C is a constant
\(\Rightarrow \smallint \frac{{dx}}{{x\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{1}{8}\log \left| x \right| - \frac{1}{4}\log \left| {x - 2} \right| + \frac{1}{8}\log \left| {x - 4} \right| + C\)
