[tex]\displaystyle\int(2(x+1)^{-1/2}-1)\,\mathrm dx[/tex]
Here we would substitute [tex]y=x+1[/tex], so that [tex]\mathrm dy=\mathrm dx[/tex], and so
[tex]\displaystyle\int(2(x+1)^{-1/2}-1)\,\mathrm dx=\int(2y^{-1/2}-1)\,\mathrm dy[/tex]
I don't really see a way around doing this with a substitution. This sort of process is involved with any integrand that includes a composite function.
Now you can integrate term-by-term with the power rule:
[tex]\displaystyle\int(2y^{-1/2}-1)\,\mathrm dy=\dfrac{2y^{1/2}}{\frac12}-y+C[/tex]
[tex]=4y^{1/2}-y+C[/tex]
[tex]=4(x+1)^{1/2}-(x+1)+C[/tex]
[tex]=4(x+1)^{1/2}-x+C[/tex]
