Mathematics

Solve :
$$ I = \int \dfrac {e^x}{ e^{2x} - 3e^x + 1 } dx $$


SOLUTION
$$\begin{array}{l} \int { \dfrac { { { e^{ x } } } }{ { { e^{ 2x } }-3{ e^{ x } }+1 } }  }  \\ Let\, { e^{ x } }=t\Rightarrow { e^{ x } }\dfrac { { dt } }{ { dx } }  \\ \int { \dfrac { { dt } }{ { { t^{ 2 } }-3t+1 } }  } =\int { \dfrac { { dt } }{ { { { \left( { t-\dfrac { 3 }{ 2 }  } \right)  }^{ 2 } }-\dfrac { 5 }{ 4 }  } }  }  \\ =\int { \dfrac { { dt } }{ { { { \left( { t-\dfrac { 3 }{ 2 }  } \right)  }^{ 2 } }-{ { \left( { \dfrac { { \sqrt { 5 }  } }{ 2 }  } \right)  }^{ 2 } } } }  }  \\ =\dfrac { 1 }{ { \sqrt { 5 }  } } \ln { \left( { \dfrac { { t-\dfrac { 3 }{ 2 } -\dfrac { { \sqrt { 5 }  } }{ 2 }  } }{ { t-\dfrac { 3 }{ 2 } +\dfrac { { \sqrt { 5 }  } }{ 2 }  } }  } \right)  } +C \\ =\dfrac { 1 }{ { \sqrt { 5 }  } } \ln { \left( { \dfrac { { 2{ e^{ x } }-3-\sqrt { 5 }  } }{ { 2{ e^{ x } }-3+\sqrt { 5 }  } }  } \right)  } +C \end{array}$$
Hence,
solved.
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Subjective Medium Published on 17th 09, 2020
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