Mathematics

$$\displaystyle \int{\dfrac{1-x^{2}}{(1+x^{2})\sqrt{1+x^{4}}}dx}$$ is equal to 


ANSWER

$$\dfrac{1}{\sqrt{2}}\sin^{-1}\left\{\dfrac{\sqrt{2}x}{x^{2}+1}\right\}+c$$


SOLUTION
$$\begin{array}{l} \int { \frac { { 1+{ x^{ 2 } } } }{ { \left( { 1+{ x^{ 2 } } } \right) \sqrt { 1+{ x^{ 4 } } }  } } dx }  \\ \int { \frac { { \left( { 1-{ x^{ 2 } } } \right)  } }{ { { x^{ 2 } }\left( { x+\frac { 1 }{ x }  } \right) \sqrt { { x^{ 2 } }+\frac { 1 }{ { { x^{ 2 } } } }  }  } } dx }  \\ =\int { \frac { { \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) dx } }{ { \left( { x+\frac { 1 }{ x }  } \right) \sqrt { { x^{ 2 } }+\frac { 1 }{ { { x^{ 2 } } } }  }  } }  }  \\ =-\int { \frac { { \left( { 1-\frac { 1 }{ { { x^{ 2 } } } }  } \right) dx } }{ { \left( { x+\frac { 1 }{ x }  } \right) \sqrt { { { \left( { x+\frac { 1 }{ x }  } \right)  }^{ 2 } }-2 }  } }  }  \\ x+\frac { 1 }{ x } =t \\ \frac { { dt } }{ { dx } } =1-\frac { 1 }{ { { x^{ 2 } } } }  \\ =-\int { \frac { { dt } }{ { t\sqrt { { t^{ 2 } }-1 }  } }  }  \\ =-\frac { 1 }{ { \sqrt { 2 }  } } { \sec ^{ -1 }  }\left( { \frac { t }{ { \sqrt { 2 }  } }  } \right) +C \\ =-\frac { 1 }{ { \sqrt { 2 }  } } \left[ { \frac { \pi  }{ 2 } -\cos  e{ c^{ -1 } }\frac { 1 }{ { \sqrt { 2 }  } }  } \right] +C \\ =\frac { 1 }{ { \sqrt { 2 }  } } \cos  e{ c^{ -1 } }\left( { \frac { 1 }{ { \sqrt { 2 }  } }  } \right) +C' \\ =\frac { 1 }{ { \sqrt { 2 }  } } { \sin ^{ -1 }  }\left( { \frac { { \sqrt { 2 }  } }{ t }  } \right) +C' \\ =\frac { 1 }{ { \sqrt { 2 }  } } { \sin ^{ -1 }  }\left( { \frac { { \sqrt { 2 }  } }{ { x+\frac { 1 }{ x }  } }  } \right) +C' \\ =\frac { 1 }{ { \sqrt { 2 }  } } { \sin ^{ -1 }  }\left( { \frac { { \sqrt { 2 } x } }{ { { x^{ 2 } }+1 } }  } \right) +C' \\ Hence,\, option\, B\, is\, the\, correct\, answer. \end{array}$$
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Single Correct Medium Published on 17th 09, 2020
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