The method step by step

Let's apply the complete method to the example $\displaystyle\int 2x\,e^{x^2}\,\mathrm{d}x$. Four steps, always the same.

Step	On this example
1. Set	$u = x^2$
2. Differentiate	$\mathrm{d}u = 2x\,\mathrm{d}x$
3. Substitute	$\displaystyle\int 2x\,e^{x^2}\,\mathrm{d}x = \int e^u\,\mathrm{d}u$
4. Integrate, then go back to $x$	$\displaystyle\int e^u\,\mathrm{d}u = e^u + C = e^{x^2} + C$

Step 2 is the key. Differentiating $u = x^2$ gives $\dfrac{\mathrm{d}u}{\mathrm{d}x} = 2x$, which we rearrange as $\mathrm{d}u = 2x\,\mathrm{d}x$. This little block $2x\,\mathrm{d}x$ is exactly what appears in the integral: we replace it in one go with $\mathrm{d}u$. It is this substitution that makes the $x$ disappear.

Be careful with step 4: after integrating in $u$, you must go back to the original variable $x$ by replacing $u$ with its expression. An antiderivative must be expressed in terms of $x$, not $u$.

What if the factor is not exactly $u'$? We adjust by a constant. For $\displaystyle\int x\,e^{x^2}\,\mathrm{d}x$, the factor $2$ is missing: we write $x\,\mathrm{d}x = \tfrac{1}{2}\left(2x\,\mathrm{d}x\right) = \tfrac{1}{2}\,\mathrm{d}u$.

$$\int x\,e^{x^2}\,\mathrm{d}x = \frac{1}{2}\int e^u\,\mathrm{d}u = \frac{1}{2}e^{x^2} + C$$

Remember the sequence: set $u$, derive $\mathrm{d}u$ from it, substitute, integrate, go back to $x$. It is mechanical once the right $u$ has been spotted.