Use substitution of $x =\tan\theta$ to show that Use the substitution $x = \tan\theta$ to show that $$\int\frac {1 - x^2} {(1+x^2)^2} dx = \int \cos2\theta\ d\theta $$
I'm a bit lost on how to handle this question, I have tried subtituting $d\theta/dx = 1 / \sec^2\theta$ but I still don't reach the answer.

I have done with your substitution yet it is so boring
$$\int  \frac { 1-x^{ 2 } }{ (1+x^{ 2 }) } dx=\int { \frac { 1-\tan ^{ 2 }{ \theta  }  }{ 1+\tan ^{ 2 }{ \theta  }  } \frac { 1 }{ \cos ^{ 2 }{ \theta  }  } d\theta =\int { \left( 1-\tan ^{ 2 }{ \theta  }  \right) d\theta =\int { \frac { \cos ^{ 2 }{ \theta -\sin ^{ 2 }{ \theta  }  }  }{ \cos ^{ 2 }{ \theta  }  }  } d\theta =\int { \frac { \cos { 2\theta  }  }{ \cos ^{ 2 }{ \theta  }  } d\theta  } = }  } \\ =\int { \cos { 2\theta \quad d\left( \tan { \theta  }  \right)  }  } $$ 
the last part we can evaluate by parts 

$$\\ \\ \int { \cos { 2\theta \quad d\left( \tan { \theta  }  \right)  }  } =\cos { 2\theta \tan { \theta +2\int { \sin { 2\theta  } \cdot \tan { \theta = }  }  }  } \cos { 2\theta \tan { \theta +4\int { \sin ^{ 2 }{ \theta d\theta  }  }  }  } =\cos { 2\theta \tan { \theta +2\int { \left( 1-\cos { 2\theta  }  \right) d\theta  } = }  } \\ =\cos { 2\theta \tan { \theta +2\left( \theta -\frac { \sin { 2\theta  }  }{ 2 }  \right)  }  } =\cos { 2\theta \tan { \theta +2\theta -\sin { 2\theta +C }  }  } $$

