Mathematics

If $$\displaystyle f\left( x \right)=\int { \frac { { x }^{ 2 }dx }{ \left( 1+{ x }^{ 2 } \right) \left( 1+\sqrt { 1+{ x }^{ 2 } }  \right)  }  } $$ and $$f\left( 0 \right) =0$$, then the value of $$f(1)$$ is


ANSWER

$$\displaystyle \log { \left( 1+\sqrt { 2 }  \right)  } -\frac { \pi  }{ 4 } $$


SOLUTION
$$\displaystyle f\left( x \right) =\int { \frac { { x }^{ 2 }dx }{ \left( 1+{ x }^{ 2 } \right) \left( 1+\sqrt { 1+{ x }^{ 2 } }  \right)  }  } $$
Substitute $$x=\tan { \theta  } \Rightarrow dx=\sec ^{ 2 }{ \theta  } d\theta =\left( 1+{ x }^{ 2 } \right) d\theta $$
$$\displaystyle \therefore f\left( x \right) =\int { \frac { { x }^{ 2 }dx }{ \left( 1+{ x }^{ 2 } \right) \left( 1+\sqrt { 1+{ x }^{ 2 } }  \right)  }  } =\int { \frac { \tan ^{ 2 }{ \theta  } \sec ^{ 2 }{ \theta  } d\theta  }{ \sec ^{ 2 }{ \theta  } \left( 1+\sec { \theta  }  \right)  }  } $$
$$\displaystyle =\int { \frac { \tan ^{ 2 }{ \theta  } d\theta  }{ 1+\sec { \theta  }  }  } =\int { \frac { \sin ^{ 2 }{ \theta  } d\theta  }{ \cos { \theta  } \left( 1+\cos { \theta  }  \right)  }  } =\int { \frac { 1-\cos ^{ 2 }{ \theta  } d\theta  }{ \cos { \theta  } \left( 1+\cos { \theta  }  \right)  }  } $$
$$\displaystyle =\int { \frac { \left( 1-\cos { \theta  }  \right)  }{ \cos { \theta  }  } d\theta  } =\int { \sec { \theta  } d\theta  } -\int { d\theta  } \\ =\log { \left( x+\sqrt { 1+{ x }^{ 2 } }  \right)  } -\tan ^{ -1 }{ x } +c\\ f\left( 0 \right) =\log { \left( 0+\sqrt { 1+0 }  \right)  } -\tan ^{ -1 }{ 0 } +c\\ \Rightarrow 0=\log { 1 } -0+c\Rightarrow c=0$$
$$\displaystyle \therefore f\left( 1 \right) =\log { \left( 1+\sqrt { 1+{ 1 }^{ 2 } }  \right)  } -\tan ^{ -1 }{ 1 } =\log { \left( 1+\sqrt { 2 }  \right)  } -\frac { \pi  }{ 4 } $$
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Single Correct Medium Published on 17th 09, 2020
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