Integrate $\int\frac{x+1}{\sqrt{1-x^2}} \; dx$ without using trigonometric substitution I want to solve: 

$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$

I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?

$$\int  \frac { x+1 }{ \sqrt { 1-x^{ 2 } }  } dx=-\left( \int  \frac { -x }{ \sqrt { 1-x^{ 2 } }  } dx-\int { \frac { dx }{ \sqrt { 1-x^{ 2 } }  }  }  \right) =\\ =-\frac { 1 }{ 2 } \int  \frac { d\left( 1-{ x }^{ 2 } \right)  }{ \sqrt { 1-x^{ 2 } }  } +\int { \frac { dx }{ \sqrt { 1-x^{ 2 } }  } =-\sqrt { 1-x^{ 2 } } +\arcsin { x } +C } $$
