jmc

For $(ax+b)(x^5+1)-(5x+1)$ to be divisible by $x^2+1,$ it must be equal to 0 at the roots of $x^2+1=0,$ which are $\pm i.$

For $x=i,$ $$\begin{array}{rl}(ax+b)(x^5+1)-(5x+1)& =(ai+b)(i+1)-(5i+1)\\ & =-a+ai+bi+b-5i-1\\ & =(-a+b-1)+(a+b-5)i.\end{array}$$Then we must have $-a+b-1=a+b-5=0.$ Solving, we find $(a,b)=\boxed{(2,3)}.$

For these values, $$(ax+b)(x^5+1)-(5x+1)=2x^6+3x^5-3x+2=(x^2+1)(2x^4+3x^3-2x^2-3x+2).$$