Mathematics

$$\int { { e }^{ x } } { \left( \dfrac { x-1 }{ { x }^{ 2 }+1 }  \right)  }^{ 2 }dx$$ 


ANSWER

$$\dfrac { { e }^{ x } }{ { x }^{ 2 }+1 } +c$$


SOLUTION
$${\int {{e^x}\left( {\dfrac{{x - 1}}{{{x^2} + 1}}} \right)} ^2}dx$$
$$ = \int {{e^x}\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx$$
$$ = \int {{e^2}\left[ {\dfrac{{\left( {{x^2} + 1} \right) - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]} $$
$$ = \int {{e^x}} \left[ {\dfrac{{{x^2} + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} - \dfrac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]$$
$$ = \int {{e^x}} \left[ {\dfrac{1}{{{{\left( {{x^2} + 1} \right)}}}} - \dfrac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]$$
$$ = \dfrac{{{e^x}}}{{{x^2} + 1}} + c$$
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Single Correct Medium Published on 17th 09, 2020
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