So I feel a bit strange asking a Calculus question, but this came up today while teaching.
One can check that if you start with some integral, which can be see as an "obvious u-substitution problem", that you can instead use integration by parts, and wind up with the scenario where you have have the original integral on both sides of your equation so you solve for the integral.
Example: Given $I=\int g^n(x)g'(x)dx$ we can clearly use u-substitution, but if we use integration by parts we get the equation $I=-nI+g^{n+1}$. This is nothing exciting or surprising, but it yields the observation that u-sub leads to one of these int by parts equations.
Question Is the opposite true?
What I mean to say is, if you do integration by parts and you wind up with an equation of this type, does it mean that you could of used some very clever u-substitution?
I feel like I should know this, but I have thought about it today, and asked a friend or two, and we don't see an immediate proof of this.
Thanks!
Best Answer
Added: I interpreted the question as follows: suppose that we use the integration by parts formula, $$\int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx$$ and $\int v(x)u'(x)\,dx = \alpha \int u(x)v'(x)\,dx$ for some constant $\alpha$, so that we can "solve" for the original integral as $$\int u(x)v'(x)\,dx = \frac{1}{1+\alpha}u(x)v(x) + C.$$ Is it the case that we can solve the original integral by doing a direct substitution instead of integration by parts?
I think this works:
Suppose that $v(x)u'(x) = \alpha u(x)v'(x)$ for some $\alpha\neq -1$. (I think this is essentially what you have, since the integral on the left hand side is $\int u(x)v'(x)\,dx$, and the integral on the right is $\int v(x)u'(x)\,dx$). I restrict to $\alpha\neq -1$, because if $\alpha=-1$, then you cannot "solve" for the original integral. But in fact, from the work below we will get that $u=\frac{B}{|v|}$, so that integration by parts will simply result in the true (but useless) $\int u\,dv = B + \int u\,dv$.
If $u$ or $v$ are zero, then the original integral was the integral of $0$, so we may discard that case. So we get that $\frac{u'}{u} = \alpha \frac{v'}{v}$, and integrating we get $\ln|u|=\alpha\ln|v|+C$, or $|u| = A|v|^{\alpha}$ for some $A\gt 0$; hence $u=B|v|^{\alpha}$ for some $B\neq 0$. So the original integral was $$\int u(x)v'(x)\,dx = \int B|v(x)|^{\alpha}v'(x)\,dx$$ which suggests the substitution $g=v(x)$ to get $B\int |g|^{\alpha}\,dg$.
Added: Note that if $\alpha = -1$, we still get $u=B|v|^{-1}$, and so again the substitution $g=|v(x)|$ will do the trick.