Substituting $x$ with another variable having an even index

Question

Often we substute $x$ with another variable having even/odd indices in order to get integral way easier and shorter than usual route.

Lets say for an function $\frac1 {(x-3)\sqrt{x+1}}$ we considered $x+1=z^2$. If we use this to get the integral then we might obtain integral=$\frac1 2ln|\frac{z-2} {z+2}|+c$. Now what is the value of $z$? Is it $z=\sqrt{x+1}$ or $z=\pm\sqrt{x+1}$?

I have faced aame sort of confusion for a lot of problems like this. However my textbook ignores the $"-"$ sign. I don't know if it is correct or not. Getting rid of thos confusion will help me a lot since this has been troubling while doing most of the problems my textbook contains.

I have looked in wolframalpha for this form of integral but found out that the site has solved the same function but in completely different ways which I am unfamiliar with.

Apparently it looks there should be $"\pm"$ but I am not sure whether my considerations are all right or not.

Also consider functions like $\frac 1 {(ax+b)(cx^n+d)}$

Another example:

$\int{x\sqrt{4-x} dx}$

If we consider $4-x=z^2$ then what should be the value of z?

Answer 1

Observe that $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$ is not bijective and so, it does not have an inverse.

However, $f:\mathbb{R}_+\to \mathbb{R}_+$ given by $f(x)=x^2$ is bijective. Thus,, in general, when you have $x=z^2$, it would mean $z=|x|$.

Answer 2

Let's look at the example $\int x\sqrt{4-x}\,dx$, $4-x=z^2$. We get $x=4-z^2$, $dx=-2z\,dz$. If we take $z>0$, then $\sqrt{4-x}=z$, and $$\int x\sqrt{4-x}\,dx=2\int(z^4-4z^2)\,dz={2z^5\over5}-{8z^3\over3}+C={2(4-x)^{5/2}\over5}-{8(4-x)^{3/2}\over3}+C$$ If we take $z<0$, then $\sqrt{4-x}=-z$, and $$\int x\sqrt{4-x}\,dx=2\int(4z^2-z^4)\,dz={8z^3\over3}-{2z^5\over5}+C=-{8(4-x)^{3/2}\over3}+{2(4-x)^{5/2}\over5}+C$$ So you get the same answer, either way. It doesn't matter whether you take $z=\sqrt{4-x}$ or $z=-\sqrt{4-x}$.