Integrating $\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx$ using two substitutions $z = x^2$ and $u = 2z-1$

Question

An exercise asks to calculate $$\int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx$$

The solution manual states that with the two substitutions $z = x^2$ and then $u = 2z -1$ we get $$\int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx = \int_1^{y^2}\frac{1}{2\sqrt{2z-1}}dz = \int_1^{2y^2-1}\frac{1}{4\sqrt{u}}du$$ I don't understand the first substitution. $\frac{dz}{dx} = 2x$, but I just don't get how the root in the denominator was transformed! Shouldn't $2x$ be a factor in the integrand for this substitution to work?

Answer 1

$$z = x² \rightarrow dz = 2 x dx \rightarrow \frac{dz}{dx} = 2x$$ $$dx = \frac{dz}{2x} \rightarrow \int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx = \int_1^{y²}\frac{1}{\sqrt{2-\frac{1}{x^2}}} \frac{dz}{2x} = \int_1^{y²}\frac{1}{\sqrt{8x²-4}} dz = \int_1^{y²}\frac{1}{\sqrt{8z-4}} dz = \int_1^{y^2}\frac{1}{2\sqrt{2z-1}}dz$$

The top boundary changes because you are in the $z$ world in which $z = x²$ so $y \rightarrow y²$.

Then with $u= 2z-1$ I assume you've understood.

Answer 2

Since one of the bounds of the integral is $1$ and you can't go past $0$, you can assume $x>0$. The substitution $x=\sqrt{z}$ thus yields $$ \sqrt{2-\dfrac{1}{x^2}}=\frac{\sqrt{2z-1}}{\sqrt{z}} $$ and $$ dx=\dfrac{1}{2\sqrt{z}}\,dz $$ so the integral becomes $$ \int_1^{y^2}\frac{\sqrt{z}}{\sqrt{2z-1}}\frac{1}{2\sqrt{z}}\,dz =\int_1^{y^2}\frac{1}{2\sqrt{2z-1}}\,dz $$ as stated.