Supposed method of integration: “long dividing” by $d$

I came across this seemingly interesting, but poorly exemplified, method of integrating:

Since integration is the inverse of differentiation, you can think of integration as “dividing” by $d$.

J. P. Ballantine [1] shows that you can formally divide by $d$ and get the correct integral. For example, he arrives at $$\int x^2 \sin x\,dx=(2-x^2)\cos x + 2x\sin x + C$$ using long division!

[1] J. P. Ballantine. Integration by Long Division. The American Mathematical Monthly, Vol. 58, No. 2 (Feb., 1951), pp. 104-105

Surely the claim that this is rigorous is questionable. However, it would be fascinating to understand how to follow this method and maybe even understand why it works; the example has no explanation, and I can’t put the steps together via reverse engineering.

Could someone shed some light on this technique?

addendum: An additional example would really complete the perfect response $\ddot\smile$

Solution 1:

I managed to "reverse engineer" the steps and I am going to go through them for you!

First, write the function you wish to integrate on the left

$$x^2 \sin x\ dx$$

Now find a function such that, when you differentiate it, the $x^2 \sin x$ comes up. Ah! When you differentiate $-x^2 \cos x$ you get $x^2 \sin x - 2x \cos x$, so it has the bit you want and the other bit seems "smaller" in the sense of "degree" or something. Just write that $-x^2 \cos x$ on the side, differentiate it and subtract the derivative, getting:

Now it's like we found an anti-derivative for the function we want to integrate, but in doing so we got the extra term $2x \cos x$ that we need to cancel... Think of something that, when differentiated, arrives at $-2x \cos x$... Ah! When you differentiate $2x \sin x$, you get the $2x \cos x$ and an extra $2 \sin x$. Write the $2x \sin x$ on the right, differentiate it, and subtract from whatever you had, getting:

and you keep going. The point being, you are always trying to come up with something (in some sense, smaller) that counters the leftovers you have from finding the anti-derivatives of the other terms!

Was I clear enough?

Let us integrate $x \log x$ just for the fun of it. Can you come up with something that, when differentiated, gives $x \log x + $ something? Of course you can, just notice that $(\frac12x^2\log x)' = x\log x + \frac12 x$

So we now have to deal with the extra $-\frac12 x$.

Can you come up with something that cancels that $-\frac12 x$? Of course you can, because $-\frac14 x^2$ does it! We now get

$$0$$

meaning we are done and

$$\int x \log x = \frac12 x^2 \log x - \frac14 x^2 + C$$