Correct Answer - Option 2 :
\(\frac{3}{5}\log \left| {x + 2} \right| + \frac{7}{5}\log \left| {x - 3} \right| + C\)
Concept:
Partial Fraction:
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 \(\smallint \frac{{2x + 1}}{{\left( {x + 2} \right)\;\left( {x - 3} \right)}}dx\)
Let \(\frac{{2x + 1}}{{\left( {x + 2} \right)\;\left( {x - 3} \right)}} = \frac{A}{{\left( {x + 2} \right)}} + \frac{B}{{\left( {x - 3} \right)}}\)
⇒ 2x + 1 = A (x - 3) + B (x + 2) ----(1)
By putting t = - 2 on both the sides of (1) we get A = 3/5
By putting t = 3 on both the sides of (1) we get B = 7/5
\(\Rightarrow \frac{{2x + 1}}{{\left( {x + 2} \right)\;\left( {x - 3} \right)}} = \frac{3}{{5\left( {x + 2} \right)}} + \frac{7}{{5\left( {x - 3} \right)}}\)
\(\Rightarrow \smallint \frac{{2x + 1}}{{\left( {x + 2} \right)\;\left( {x - 3} \right)}}dx = \frac{3}{5}\;\smallint \frac{{dx}}{{x + 2}} + \frac{7}{5}\;\smallint \frac{{dx}}{{x - 3}}\)
As we know that \(\smallint \frac{{dx}}{x} = \log \left| x \right|\; + C\) where C is a constant
\(\Rightarrow \smallint \frac{{2x + 1}}{{\left( {x + 2} \right)\;\left( {x - 3} \right)}}dx = \frac{3}{5}\log \left| {x + 2} \right| + \frac{7}{5}\log \left| {x - 3} \right| + C\)