solve a integral without using without using trigo-substitution?
I think it's an useless question, 'cause it's an immediate integral.
But you can say $\displaystyle y=\arctan x\implies\tan y=x\,\therefore\,y'(1+x^2)=1.$
So, $\displaystyle y'=\frac1{1+x^2}.$
Now integrate both sides $\displaystyle y+k=\int\frac1{1+x^2}\,dx.$
Since $\displaystyle y=\arctan x,$ we happily get that $\displaystyle \arctan x+k=\int\frac1{1+x^2}\,dx.$