Thomas' Calculus 13th Edition

$(f\circ g\circ h)(x)=\sqrt{\dfrac{5x+1}{4x+1}}.$
We are given $f(x)=\sqrt{x+1}$, $g(x)=\dfrac{1}{x+4}$, and $h(x)=\dfrac{1}{x}$. So $(f\circ g\circ h)(x)=f(g(h(x)))=f\left(g\left(\dfrac{1}{x}\right)\right)=f\left(\dfrac{1}{\dfrac{1}{x}+4}\right)=\sqrt{\dfrac{1}{\dfrac{1}{x}+4}+1}$. Now, we simplify. We first multiply the numerator and denominator of our fraction by $x$ to get $\sqrt{\dfrac{x}{1+4x}+1}.$ Next, we find a common denominator for the expression under the radical sign, which gives $\sqrt{\dfrac{x}{1+4x}+\dfrac{1+4x}{1+4x}}=\sqrt{\dfrac{x+1+4x}{1+4x}}=\sqrt{\dfrac{5x+1}{1+4x}}=\sqrt{\dfrac{5x+1}{4x+1}}.$