For (21) Let \displaystyle \color{blue}I =\int\frac{x-1}{(x+x\sqrt{x}+\sqrt{x})\sqrt{\sqrt{x}(x+1)}}dx.

Now put x=t^2 and dx=2tdt

So we have \displaystyle I =2\int\frac{t^2-1}{(t^2+t+1)\sqrt{t^3+t}}dx

\displaystyle \Rightarrow I =2\int\frac{1-\frac{1}{t^2}}{\bigg(t+\frac{1}{t}+1\bigg)\sqrt{t+\frac{1}{t}}}dt.

Now put \displaystyle t+\frac{1}{t}=u^2. Then \displaystyle \bigg(1-\frac{1}{t^2}\bigg)dt=2udu.

So We have \displaystyle I =4\int\frac{1}{u^2+1}du=4\arctan(u)+\mathcal{C}.

So finally we get \displaystyle \color{brown}I =4\arctan\bigg(\sqrt{\frac{x+1}{\sqrt{x}}}\bigg)+\mathcal{C}.