# Constants of integration in integration by parts

After finishing a first calculus course, I know how to integrate by parts, for example, $\int x \ln x dx$, letting $u = \ln x$, $dv = x dx$: $$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$

However, what I could not figure out is why we assume from $dv = x dx$ that $v = \frac{x^2}{2}$, when it could be $v = \frac{x^2}{2} + C$ for any constant $C$. The second integral would be quite different, and not only by a constant, so I would like to understand why we "forget" this constant of integration.

Thanks.

The second integral would change, but also the first term... Have you actually checked to see what happens if you change the constant?

Take your example, $$\int x\ln x\,dx.$$ Note $x\gt 0$ must be assumed (so the integrand makes sense).

If we let $u = \ln x$ and $dv= x\,dx$, then we can take $v$ to be any function with $dv = x\,dx$. So the "generic" $v$ will be, as you note, $v = \frac{1}{2}x^2 + C$. What happens then if we use this "generic" $v$? \begin{align*} \int x\ln x\,dx &= \ln x\left(\frac{1}{2}x^2 + C\right) - \int \left(\frac{1}{2}x^2+C\right)\frac{1}{x}\,dx\\ &= \frac{1}{2}x^2\ln x + C\ln x - \int\left(\frac{1}{2}x + \frac{C}{x}\right)\,dx\\ &= \frac{1}{2}x^2\ln x + C\ln x - \frac{1}{4}x^2 - C\ln x + D\\ &= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + D, \end{align*} so in the end, we get the same result no matter what value of $C$ we take for $v$.

This says that we can take any value of $C$ and still get the same answer. Since we can take any value of $C$, why not take the simplest one, the one that does not require us to carry around an extra term that is going to cancel out anyway? Say..., $C=0$?

This works in general. If you replace $v$ with $v+C$ in the integration by parts formula, you have \begin{align*} \int u\,dv &= u(v+C) - \int(v+C)\,du = uv + Cu - \int v\,du - \int C\,du\\ &= uv+Cu - \int v\,du - Cu = uv-\int v\,du. \end{align*} So the answer is the same regardless of the value of $C$, and so we take $C=0$ because that makes our life simpler.

• I think you've proved that $C=0$ in this case. We're not just making an assumption here. $C=0$ is a hard concrete result in the by-parts formula. – Nick Nov 1 '14 at 9:44

Your observation that $dv=xdx$ does not imlpy $v=x^2/2$ is correct.

Your confusion resolves when you say it this way: we set $v=x^2/2$ and this implies $dv=xdx$.

$$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$

You could always write $$v = \frac{x^2}{2} + C$$ but that won't matter much because the final result would also involve a constant (Say $K$ which would be equal to $C+k$ )

• This is not exactly what happens... – Mariano Suárez-Álvarez Mar 14 '11 at 7:48
• Arturo's answer makes me realize this is ok. If you make an edit, howerver small, I will remove my downvote. – Please Delete Account Mar 14 '11 at 15:33
• @Approximist : Done! – Prasoon Saurav Mar 14 '11 at 16:44

HINT $\rm\ \ C'=0\ \ \Rightarrow\ \ (UV)'-U'\:V\ =\ UV'\: =\ U(V+C)'\: =\ (U(V+C))'-U'\:(V+C)$

We didn't "forget". We simply choose C in a way that the resulting $\int u\,dv$ would be the simplest form. Usually $C = 0$ but not always. If, for example, we have this integral:

$$\int \ln(x+2) \,dx$$

Then you would choose $v = x + 2$ because $du = \frac{dx}{x+2}$

Second example:

$$\int x\ln(x+2) \,dx$$

Then $$v = \frac{x^2-4}{2} = \frac{(x-2)(x+2)}{2}$$ and $$u\,dv = \frac{x-2}{2} dx$$

We "forget" it, and add it in the last step. The whole point in the constant of integration is to remind us that there could have been a constant term added on at the end originally, but in the process of differentiation we got rid of it because it did not affect the slope.