Differentials can be treated as a fraction in the sense you think of it; however it needs some proofs using limits to show that they can be. However, when treating differentials as fractions (for example cancelling terms out) you think of "dx" as an infinitesimal quantity, you can't think x as a normal variable and d a very small real which makes the product infinitesimal. "d" alone means nothing, so you can't conclude that "dy/dx" is "y/x" by cancelling d's out.
Solving differential equations relies on the chain rule to establish that the derivative holds some of the properties a fraction does (not all!). The method you are talking about seems like u-substitution. When doing it, you merely write the differential dx in terms of du, for example, consider this integral:
\int^{1}_{0}(3x+5)^2 dx
When you substitute in u=3x+5, first you express x in terms of u: x=\dfrac{u-5}{3}. Then all you do is to express the differential dx in terms of du. How do you do this? Consider the well-known representation of the derivative, dx/du. To find this value all we do is to differentiate the right hand side with respect to u, which yields 1/3, and hence we obtain that \dfrac{dx}{du}=\dfrac{1}{3}. Multiplying both sides by du yields (this is the step where the chain rule comes in: we treated the derivative as a fraction) dx=\dfrac{du}{3}. Substituting this in, we simply obtain
\int_{0}^{1}u^2 \dfrac{du}{3}
But wait, is that right? We carried the integral from the x-space to the u-space, but we are still using the same limits of integration! We have to change those. To do this, we simply plug in the two limits as x in the formula of u, and the upper limit becomes 8 and the lower limit becomes 5. This gives us the final integral
\frac{1}{3}\int_{5}^{8}u^2 du
whose evaluation is straightforward.
And about the integration by parts formula: Write down the product rule and take the integral of both sides.
