$$\dot{x} = \Bigg\{-t \sqrt{|x|}, \, \, x \geq 0
\, \, \, t \sqrt{|x|} , x \leq 0\Bigg\}$$

$\begin{array}{rcl}
\dfrac{\mathrm dx}{\mathrm dt} &=& x^2 t\\
\dfrac{\mathrm dx}{x^2} &=& t\ \mathrm dt\\
\displaystyle \int \dfrac{\mathrm dx}{x^2} &=& \displaystyle \int t\ \mathrm dt\\
\dfrac{-1}{x} &=& \dfrac12 t^2 + C\\
x &=& \dfrac{-2}{t^2 + 2C}\\
\end{array}$

$$\dot{x} = \Bigg\{-t \sqrt{|x|}, \, \, x \geq 0
\, \, \, t \sqrt{|x|} , x \leq 0\Bigg\}$$


My intial appoarch was to consider if $x \geq \, 0$ then we could rewrite as follows:


$$\frac{dy}{dx} = -t \sqrt{|x|}dt$$

Then I considered if $x \, \leq 0$ we have the following:
$$\frac{dy}{dx} = t \sqrt{|x|}dt$$

$\begin{array}{rcl}
\dfrac{\mathrm dx}{\mathrm dt} &=& -t \sqrt{x} \\
\dfrac{\mathrm dx}{\sqrt x} &=& -t \ \mathrm dt \\
2\sqrt x &=& -0.5 t^2 + C \\
x &=& (-0.25 t^2 + C)^2 \\
\end{array}$

1/2i ( e^iw - e^-iw ) = i
( e^iw - e^-iw ) = 2i^2 = -2
e^2iw - 1 = -2e^iw
e^2iw + 2e^iw - 1 = 0
e^2iw + 2e^iw + 1 = 2
(e^iw+1)^2 = 2

In integration by substitution where you use a substitution in the form u^2
Why do you then multiply your substituted function by 2u

First of all, you can split it into two fractions due to the numerator, one a simple $\int\frac a{(b+cu)^{3/2}}\ du$ and the other of the form $\int\frac{au}{(b+cu)^{3/2}}\ du$. The second is handled as follows:
Consider the easier integral:
$$\int\frac1{(a+bu)^{1/2}}\ du$$
Solve, then differentiate with respect to $b$.
@Lozansky

@sbm
Once you add it as a bookmark then how do you use it?

$$\int {\frac {x}{\sqrt{1+x}}} dx $$
using $$u^2=1+x $$

When we change it to $$ \int{\frac{u^2-1}{u} 2u} du $$

Where is the du/dx since we are using the fact that $$\int{f(u) \frac{du}{dx} dx = \int f(u) du $$

$\int\sqrt{1+xt}\ dx=\frac2{3t}(1+xt)^{3/2}$
Differentiate both sides with respect to $t$ and set $t=1$.

For example $ \mathrm dx = 2u \ \mathrm du $
I can see how it works but it wouldn't be intuitive as $2u du$ is meaningless

@Lozansky I get this:
$$\int\frac u{-2(a+bu)^{3/2}}\ du\frac\partial{\partial b}\int\frac1{(a+bu)^{1/2}}\ du=\frac\partial{\partial b}\frac{2\sqrt{a+bu}}b=\frac u{b\sqrt{a+bu}}-\frac{2\sqrt{a+bu}}{b^2}l$$

$$\int {cosec^2x} {cot^2x} {dx} $$

Using $u=cotx$