Integration by parts as a choice, how to make it, and the two integrals that come back to where they started.
∫ u dvdx dx = uv − ∫ v dudx dx
You are trading one integral for another, and the trade is only worth making if the new one is simpler. That is the entire decision.
Take ∫ x sin x dx. Choose which part is u and watch what the second integral turns into.
One of these two choices finishes. The other spirals.
Choose u to be the thing that gets simpler when differentiated. A power of x becomes a lower power and eventually disappears. A sine or an exponential never does. So x is almost always u, and the trigonometric or exponential part is almost always dv.
The exception is ln x. It has no simple integral, so it must be u, with dv = dx. That is how ∫ ln x dx = x ln x − x + C comes out, and it is the one case where the power of x is not u.
A power of n needs n applications. Each one drops the power by one, so you can see the end from the start.
∫ exsin x dx comes back to itself. Apply parts twice and the original integral reappears on the right. That is not failure: call it I, rearrange, and solve for I. Recognising that is the trick the question is testing.
In an examination a substitution will be provided whenever the integral is not already of the reverse chain rule form. When one is given, use it: changing the limits as well as the variable is usually cleaner than substituting back.
1. For ∫ x cos x dx, which choice of u works?
2. ∫ ln x dx is:
3. Evaluate ∫01 xex dx.
Stating u and dv/dx explicitly, and then du/dx and v. Four short lines, and they carry the method marks even if the arithmetic afterwards slips.
Applying parts the right number of times, and spotting when the original integral has reappeared so you can solve for it.
On a definite integral, the uv part is evaluated at the limits too. Forgetting to apply the limits to that term is a common and expensive slip.
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