The inverse of the functiony=loga(x+(x2+1);(a>0,a≠1)  is

∵  y=loga(x+(x2+1))=f(x)(say) ∴  x=f−1(y) And ay=x+(x2+1)     …….(i)Or    a−y=(x2+1)−x ……..(ii)Subtracting  Eq. (ii) from Eq.(i),thenx=12(ay−a−y) ⇒f−1(y)=12(ay−a−y) Hence, f−1(x)=12(ax−a−x)