Good Question 12 – Parts with a Constant?

partsSomeone asked me about this a while ago and I thought I would share it with you. It may be a good question to get your students thinking about; see if they can give a definitive answer that will, of course, include a justification.

Integration by Parts is summarized in the equation

\displaystyle \int_{{}}^{{}}{udv}=uv-\int_{{}}^{{}}{vdu}

To use the equation, you choose part of a given integral (left side) to be u and part to be dv, both functions of x. Then you differentiate u and integrate dv and use them on the right side to obtain a simpler integral that you can integrate.

The question is this: When you integrate dv, should you, can you, have a constant of integration, the “+ ” that you insist upon in other integration problems? Why don’t you use it here? Or can you?

Scroll down for my answer.

Answer: Let’s see what happens if we use a constant. Assume that \displaystyle \int_{{}}^{{}}{dv}=v+C. Then

\displaystyle \int_{{}}^{{}}{udv}=u\cdot \left( v+C \right)-\int_{{}}^{{}}{\left( v+C \right)du}

\displaystyle =uv+Cu-\left( \int_{{}}^{{}}{vdu}+\int_{{}}^{{}}{Cdu} \right)

\displaystyle =uv+Cu-\int_{{}}^{{}}{vdu}-Cu

\displaystyle =uv-\int_{{}}^{{}}{vdu}

So, you may use a constant if you want, but it will always add out of the expression.

For more on integration by parts see here for the basic idea, here for the tabular method, here for a quicker way than the tabular method, and here for more on the tabular method and reduction formulas.


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