# All possible antiderivatives for Integration by Parts

Just had a quick inquiry with regards to the formula for Integration by Parts. If I'm not mistaken, the formula states that

$$\int f'(x)g(x) = f(x)g(x)- \int f(x)g'(x)$$

However, in the case that I try to substitute an antiderivative with a valid constant, the formula does not appear to work. I attempted to use a simple product such as $$7x(x^2)$$, using $$7x$$ as $$f'(x)$$ and $$x^2$$ as $$g(x)$$ respectively. I found that using $$\frac{7}{2}x^2+5$$ for the antidervative for $$7x$$ does not work in the formula, while $$\frac{7}{2}x^2$$ without the constant does indeed work. I am sure that I am missing something, however, why do both of these solutions not work, even though both are valid antiderivatives?

• General Tip: use "$" around math expressions to enter mathjax mode. here is a full tutorial. – G. Chiusole Aug 23 '19 at 22:14 • @G. Chiusole Thanks! – D_A Aug 23 '19 at 22:21 • I mean, with polynomials, integration by parts is uninformative because you cannot use it to lower the degree. So eventually you'll end up with that$x^3\$ that never goes away. – Gae. S. Aug 23 '19 at 22:29
• I suspect you messed up the calculation. – Arturo Magidin Aug 23 '19 at 22:30
• @D_A If you deem a given answer satisfactory, please mark it as the accepted answer by clicking the tick mark to the left of the answer body (it'll turn green once you click it). This way the question will not show up in the "unanswered queue" in the future. – G. Chiusole Aug 24 '19 at 9:29

Say you pick $$f(x) = \frac{7}{2}x^2 + 5$$, $$g(x) = x^2$$, and are doing $$\int f'(x)g(x)\,dx = \int (7x)x^2\,dx = \int 7x^3\,dx = \frac{7}{4}x^4 + C.$$

Now, if you try using integration by parts with the anti-derivative you pick, you have \begin{align*} \int f'(x)g(x)\,dx &= f(x)g(x) - \int g'(x)f(x)\,dx\\ &= \left(\frac{7}{2}x^2+5\right)(x^2) - \int 2x\left(\frac{7}{2}x^2+5\right)\,dx\\ &= \frac{7}{2}x^4 + 5x^2 - \int (7x^3 + 10x)\,dx\\ &= \frac{7}{2}x^4 + 5x^2 - \left( \frac{7}{4}x^4 + 5x^2 + D\right)\\ &= \frac{7}{2}x^4 - \frac{7}{4}x^4 + 5x^2 -5x^2 - D\\ &= \frac{7}{4}x^4 - D. \end{align*}That is, the same answer, up to a constant.

So long as you use the same antiderivative in both instances of $$f(x)$$ on the right hand side, it will work out. Recall that if $$f(x)$$ is one antiderivative, then every antiderivative is of the form $$f(x)+D$$, with $$D$$ a constant. So you would get: \begin{align*} (f(x)+D)g(x) &- \int (f(x)+D)g'(x)\,dx\\&= f(x)g(x) + Dg(x) -\int f(x)g'(x)\,dx - \int Dg'(x)\,dx\\ &= f(x)g(x) + Dg(x) - \int f(x)g'(x)\,dx - D\int g'(x)\,dx\\ &= f(x)g(x) + Dg(x) - \int f(x)g'(x)\,dx - D(g(x)+E)\\ &= f(x)g(x) + Dg(x) - Dg(x) - \int f(x)g'(x)\,dx - DE\\ &= f(x)g(x) - \int f(x)g'(x)\,dx \end{align*} (because that final constant gets "absorbed" into the indefinite integral).

For indefinite integrals the rule for partial integration is

$$\int f'(x) g(x) dx = f(x) g(x) - \int f(x) g'(x) dx$$

where the equality is true up to an unspecified constant $$C$$. In your case with $$f'(x) := 7x$$ and $$g(x) := x^2$$ this means that

$$\int 7x x^2 dx = \frac{7}{2} x^2 x^2 - \int \frac{7}{2} x^2 2x dx$$

and hence

$$2 \underbrace{\int 7x^3 dx}_{=\frac{7}{4} x^4+C} = \frac{7}{2} x^4 ~~.$$

Why we only have equality up to a constant is best explained by how the formula is derived: see here.

Also, for definite integrals, we have an actual equality. Here the formula is

$$\int_b^a f'(x) g(x) dx = f(x) g(x) \vert_b^a- \int_b^a f(x) g'(x) dx$$

for appropriate $$a,b \in \mathbb{R}$$.

• Thanks so much. Why is this the case? I understand what you are saying, I just do not understand why that is the case. – D_A Aug 23 '19 at 22:28
• @D_A see the link I added. In the third line we form the antiderivative, which is only defined up to a constant. – G. Chiusole Aug 23 '19 at 22:32