Mathematics

# Solve :$\int_{}^{} {\frac{{\left( {x - 1} \right)}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}} dx$

##### SOLUTION
$\begin{array}{l} \int _{ }^{ }{ \frac { { \left( { x-1 } \right) } }{ { \left( { x+1 } \right) \left( { x-2 } \right) } } } dx \\ Let\, \frac { { \left( { x-1 } \right) } }{ { \left( { x+1 } \right) \left( { x-2 } \right) } } =\frac { A }{ { \left( { x+1 } \right) } } +\frac { B }{ { \left( { x-2 } \right) } } \\ Then,\left( { x-1 } \right) \equiv A\left( { x-2 } \right) +B\left( { x+1 } \right) ......\left( i \right) \\ putting\, x=1\, in\left( i \right) ,\, we\, get\, A=\frac { 2 }{ 3 } . \\ putting\, x=2\, in\left( i \right) ,\, we\, get\, B=\frac { 1 }{ 3 } . \\ \therefore \frac { { \left( { x-1 } \right) } }{ { \left( { x+1 } \right) \left( { x-2 } \right) } } =\frac { 2 }{ { 3\left( { x+1 } \right) } } +\frac { 1 }{ { 3\left( { x-2 } \right) } } \\ =\int _{ }^{ }{ \frac { { \left( { x-1 } \right) } }{ { \left( { x+1 } \right) \left( { x-2 } \right) } } } dx \\ =\frac { 2 }{ 3 } \int _{ }^{ }{ \frac { { dx } }{ { \left( { x+1 } \right) } } } +\int _{ }^{ }{ \frac { { dx } }{ { \left( { x-2 } \right) } } } \\ =\frac { 2 }{ 3 } \log \left| { x+1 } \right| +\frac { 1 }{ 3 } \log \left| { x-2 } \right| +C \end{array}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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#### Realted Questions

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1 Verified Answer | Published on 17th 09, 2020