Consider a variable transformation expression of the form $ v=\pm u$.
For the integration of a function $f(u)$ where $u_i$ and $u_f$ are the respective limits, we have:
$$I=\int_{u_i}^{u_f} f(u) \, du$$
I intend to determinate this integral after a variable substitution $ v=\pm u$. So, I need to compute $\left|\frac{du}{dv}\right|$ which is easy, and the new limits of integration, $v_i$ and $v_f$. My problem here, is how to make this last step? $v$ is not an injective funcion of $u$, for each value of $u$ there's two values of $v$.
Maybe your interests would be best served by an example. Start off with $$\int_0^3\left(u^2+u\right)du=\left[\frac13u^3+\frac12u^2\right]_0^3=\frac{27}2$$ Now we would like to execute the same integral, but in terms of $v=u^2-2u+2$. Solving for $u$, we find that $u=1\pm\sqrt{v-1}$ and is double valued. We can spot the singular point because $$\frac{dv}{du}=2u-2=0$$ when $u=1$. That means we have to break the integral at $u=1$: $$\int_0^3\left(u^2+u\right)du=\int_0^1\left(u^2+u\right)du+\int_1^3\left(u^2+u\right)du$$ For the first integral, $u=1-\sqrt{v-1}$, when $u=0$, $v=2$, when $u=1$, $v=1$, and $$du=-\frac12\frac{dv}{\sqrt{v-1}}$$ So $$\begin{align}\int_0^1\left(u^2+u\right)du&=\int_2^1\left\{\left(1-\sqrt{v-1}\right)^2+1-\sqrt{v-1}\right\}\left(-\frac12\frac{dv}{\sqrt{v-1}}\right)\\ &=\frac12\int_1^2\frac{1-2\sqrt{v-1}+v-1+1-\sqrt{v-1}}{\sqrt{v-1}}dv\\ &=\frac12\int_1^2\left(-3+\sqrt{v-1}+\frac2{\sqrt(v-1)}\right)dv\\ &=\frac12\left[-3v+\frac23(v-1)^{\frac32}+4(v-1)^{\frac12}\right]_1^2\\ &=\frac56\end{align}$$ Now for the second integral, $u=1+\sqrt{v-1}$, when $u=1$, $v=1$, when $u=3$, $v=5$, and $$du=\frac12\frac{dv}{\sqrt{v-1}}$$ So $$\begin{align}\int_1^3\left(u^2+u\right)du&=\int_1^5\left\{\left(1+\sqrt{v-1}\right)^2+1+\sqrt{v-1}\right\}\cdot\frac12\frac{dv}{\sqrt{v-1}}\\ &=\frac12\int_1^5\frac{1+2\sqrt{v-1}+v-1+1+\sqrt{v-1}}{\sqrt{v-1}}dv\\ &=\frac12\int_1^5\left(3+\sqrt{v-1}+\frac2{\sqrt(v-1)}\right)dv\\ &=\frac12\left[3v+\frac23(v-1)^{\frac32}+4(v-1)^{\frac12}\right]_1^5\\ &=\frac{38}3\end{align}$$ The integral after the transformation evaluates to $$\int_1^3\left(u^2+u\right)du=\frac56+\frac{38}3=\frac{27}2$$ Thanking you for your kind attention to my lesson :)