Question

Integrate (2(x+1)^-1/2) -1 dx and show me how the substitution rule works. Is it possible without using the substitution rule?

Answer 1

`\displaystyle\int(2(x+1)^(-1/2)-1)\,\mathrm dx`

Here we would substitute
`y=x+1`, so that
`\mathrm dy=\mathrm dx`, and so


`\displaystyle\int(2(x+1)^(-1/2)-1)\,\mathrm dx=\int(2y^(-1/2)-1)\,\mathrm dy`

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:


`\displaystyle\int(2y^(-1/2)-1)\,\mathrm dy=(2y^(1/2))/(\frac12)-y+C`

`=4y^(1/2)-y+C`

`=4(x+1)^(1/2)-(x+1)+C`

`=4(x+1)^(1/2)-x+C`