Mathematics

Solve :
$$\int_{}^{} {\frac{{\left( {3x - 2} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x + 3} \right)}}} dx$$


SOLUTION
$$\begin{array}{l} \int _{  }^{  }{ \frac { { \left( { 3x-2 } \right)  } }{ { { { \left( { x+1 } \right)  }^{ 2 } }\left( { x+3 } \right)  } }  } dx \\ Let\, \frac { { \left( { 3x-2 } \right)  } }{ { { { \left( { x+1 } \right)  }^{ 2 } }\left( { x+3 } \right)  } } =\frac { A }{ { \left( { x+1 } \right)  } } +\frac { B }{ { { { \left( { x+1 } \right)  }^{ 2 } } } } +\frac { C }{ { \left( { x+3 } \right)  } }  \\ \Rightarrow \left( { 3x-2 } \right) \equiv A\left( { x+1 } \right) \left( { x+3 } \right) +B\left( { x+3 } \right) +C{ \left( { x+1 } \right) ^{ 2 } }.......\left( i \right)  \\ Putting\, \, x=-13\, no\, both\, side\, of\, \left( i \right) ,\, we\, get\, C=\frac { { -11 } }{ 4 }  \\ Putting\, x=-1\, on\, both\, sides\, of\left( i \right) ,\, we\, get\, B\, =\frac { { -5 } }{ 2 }  \\ A+C=0\Rightarrow A=-C=\frac { { 11 } }{ 4 }  \\ \therefore \frac { { \left( { 3x-2 } \right)  } }{ { { { \left( { x+1 } \right)  }^{ 2 } }\left( { x+3 } \right)  } } =\frac { { 11 } }{ { 4\left( { x+1 } \right)  } } -\frac { 5 }{ { 2{ { \left( { x+1 } \right)  }^{ 2 } } } } -\frac { { 11 } }{ { 4\left( { x+3 } \right)  } }  \\ \Rightarrow \int _{  }^{  }{ \frac { { \left( { 3x-2 } \right)  } }{ { { { \left( { x+1 } \right)  }^{ 2 } }\left( { x+3 } \right)  } }  } dx \\ =\frac { { 11 } }{ 4 } \int _{  }^{  }{ \frac { { dx } }{ { \left( { x+1 } \right)  } }  } -\frac { 5 }{ 2 } \int _{  }^{  }{ \frac { 1 }{ { { { \left( { x+1 } \right)  }^{ 2 } } } }  } dx-\frac { { 11 } }{ 4 } \int _{  }^{  }{ \frac { { dx } }{ { \left( { x+3 } \right)  } }  }  \\ =\frac { { 11 } }{ 4 } .\log  \left| { \frac { { x+1 } }{ { x+3 } }  } \right| +\frac { 5 }{ { 2\left( { x+1 } \right)  } } +C \end{array}$$
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Subjective Medium Published on 17th 09, 2020
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