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?
 A: 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 $a,b \in \mathbb{R}$.
A: 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).
