Your Question : Integral of #[(√1-x²)/1+x²] dx ?#
Your Question after formation : #intcolor(violet)(sqrt(1-x²))/color(orange)((1+x²))dx #
Answer for updated question :
Let ,
#I=intsqrt(1-x²)/(1+x²)dx #
subst. #color(red)(x=sinu=>dx=cosudu#
#I=intsqrt(1-sin^2u)/(1+sin^2u)cosudu#
#=intcos^2u/(1+sin^2u)du#
#=int1/(sec^2u+tan^2u)du#
#=int1/(1+2tan^2u)du#
#=intsec^2u/(sec^2u(1+2tan^2u))du#
#=intsec^2u/((1+tan^2u)(1+2tan^2u))du#
Subst. #color(brown)(tanu=v=>sec^2u=dv#
So,
#I=int1/((1+v^2)(1+2v^2))dv#
But, #tocolor(blue)(2(1+v^2)-(1+2v^2)=2+2v^2-1-2v^2=1#
#:.I=intcolor(blue)(2(1+v^2)-(1+2v^2))/((1+v^2)(1+2v^2))dv#
#I=int2/(1+2v^2)dv-int1/(1+v^2)dv#
#I=int1/(1/2+v^2)dv-int1/(1+v^2)dv#
#I=int1/((1/sqrt2)^2+v^2)dv-int1/(1+v^2)dv#
#=>I=1/(1/sqrt2) *arctan(v/(1/sqrt2))-1/1*arc tan(v/1)+c#
#=>I=sqrt2arc tan(sqrt2v)-arc tan (v)+c#
Subst. back , #color(brown)(v=tanu#
#I=sqrt2arc tan(sqrt2tanu)-arctan(tanu)+c#
#I=sqrt2arc tan(sqrt2sinu/cosu)-u+c#
#I=sqrt2arctan(sqrt2sinu/sqrt(1-sin^2u))-u+c#
Subst. back , #color(red)(sinu=x and u=arcsinx#
#I=sqrt2arc tan((sqrt2x)/sqrt(1-x^2))-arcsinx+c#
