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## $\int {\frac{{2x – 1}}{{(x – 1)(x + 2)(x – 3)}}} dx$ ~~~~~[NCERT Exemplar Q. 38,Page 165 ]

# Let $I = \int {\frac{{(2x – 1)}}{{(x – 1)(x + 2)(x – 3)}}} dx$

Now $\frac{{2x – 1}}{{(x – 1)(x + 2)(x – 3)}} = \frac{A}{{(x – 1)}} + \frac{B}{{(x + 2)}} + \frac{C}{{(x – 3)}}$

# $ \Rightarrow $$2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)$

Let’s put $x = 3,$ then

$6 – 1 = C(3 – 1)(3 + 2)$

$ \Rightarrow $$5 = 10C \Rightarrow C = \frac{1}{2}$

# Let’s put $x = 1$, then

$2 – 1 = A(1 + 2)(1 – 3)$

$ \Rightarrow $$1 = – 6A \Rightarrow A = – \frac{1}{6}$

# Now, Let’s put $x = – 2,$ then

$ – 4 – 1 = B( – 2 – 1)( – 2 – 3)$

# $ \Rightarrow $$ – 5 = 15B \Rightarrow B = – \frac{1}{3}$

# therefore,$I = – \frac{1}{6}\int {\frac{1}{{x – 1}}} dx – \frac{1}{3}\int {\frac{1}{{x + 2}}} dx + \frac{1}{2}\int {\frac{1}{{x – 3}}} dx$

$ = – \frac{1}{6}\log |(x – 1)| – \frac{1}{3}\log |(x + 2)| + \frac{1}{2}\log |(x – 3)| + C$

$ = – \log |(x – 1){|^{1/6}} – \log |(x + 2){|^{1/3}} + \log |(x – 3){|^{1/2}} + C$

$ = \log \left| {\frac{{\sqrt {x – 3} }}{{{{(x – 1)}^{1/6}}{{(x + 2)}^{1/3}}}}} \right| + C$

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Now $\frac{{2x – 1}}{{(x – 1)(x + 2)(x – 3)}} = \frac{A}{{(x – 1)}} + \frac{B}{{(x + 2)}} + \frac{C}{{(x – 3)}}$

Let’s put $x = 3,$ then

$6 – 1 = C(3 – 1)(3 + 2)$

$ \Rightarrow $$5 = 10C \Rightarrow C = \frac{1}{2}$

$2 – 1 = A(1 + 2)(1 – 3)$

$ \Rightarrow $$1 = – 6A \Rightarrow A = – \frac{1}{6}$

$ – 4 – 1 = B( – 2 – 1)( – 2 – 3)$

$ = – \frac{1}{6}\log |(x – 1)| – \frac{1}{3}\log |(x + 2)| + \frac{1}{2}\log |(x – 3)| + C$

$ = – \log |(x – 1){|^{1/6}} – \log |(x + 2){|^{1/3}} + \log |(x – 3){|^{1/2}} + C$

$ = \log \left| {\frac{{\sqrt {x – 3} }}{{{{(x – 1)}^{1/6}}{{(x + 2)}^{1/3}}}}} \right| + C$