Please explain why this equation doesn't work with this form of U-substitution?

Question

Everywhere I read online people use trig substitution to solve it, but I want to understand why in the world this isn't the right way of doing it?

$$ f(x) = \int(x^2+y^2)^{-3/2} dx $$

let: $$ u = x^2 + y^2; \space \frac{du}{dx} = 2x;\space dx = \frac{du}{2x} $$

$$ f(x) = \frac{1}{2x}\int u^ {-3/2}du $$

$$ f(x) = \frac{1}{2x}\left( -\frac{2}{u^{1/2}} \right) + C $$

$$ f(x) = \frac{1}{2x}\left( -\frac{2}{(x^2+y^2)^{1/2}} \right) + C $$

$$ f(x) = -\frac{1}{x(x^2+y^2)^{1/2}} + C $$

The answer given by integral calculators and various proofs using trig give a result of:

$$ f(x) = =\frac{x}{y^2(x^2+y^2)^{1/2}} + C $$

Why are they coming out to be completely different?

Answer 1

You can't sneak the "${1\over 2x}$" out of the integral: even though the variable of integration is $u$ instead of $x$, since $u$ depends on $x$ the $x$ is still "bound" into the integral.

It might make things clearer to consider an example you already understand well. Take $I=\int x^2dx$, and set $u=x^2$ (yes this is stupid, but do it anyways). Then $I=\int{1\over 2x}udu$. If we pull the "${1\over 2x}$" out, we would get $${1\over 2x}({u^2\over 2}+C)={x^4\over 4x}+{C\over 2x}={x^3\over 4}+{C\over 2x}$$ which is clearly incorrect; if we keep them together, and put everything in terms of $u$, we get $$\int{1\over 2u^{1\over 2}}udu={1\over 2}\int u^{1\over 2}du={1\over 2}({2\over 3}u^{3\over 2}+C)={1\over 3}x^3+C,$$ which is correct. (Note that I'm conflating $C$ with ${1\over 2}C$, there, but that's an okay abuse of notation since $C$ denotes an arbitrary constant.)

Answer 2

$x$ is a function of $u$, not a constant, so you can't take it out of the integral. You have to turn it into an explicit function $u$: $u=x^2+y^2$, so $x=\sqrt{u-y^2}$. Thus your substitution actually turns the integral into $$ \int \frac{u^{-3/2}}{(u-y^2)^{1/2}} \, du, $$ which is not really an improvement over the original expression.