Integration by parts!
Assuming product rule for differentiation is true (here’s a link to a nice little proof for that), we can get the rule for integration by parts!
If f(x) = g(x)h(x), then f’(x) = g’(x)h(x) + g(x)h’(x) [This is just the product rule]
f(x) +c = ∫[g’(x)h(x) + g(x)h’(x)].dx [I’ve integrated both sides with respect to x indefinitely.]
f(x) +c = ∫g’(x)h(x).dx + ∫g(x)h’(x).dx [I’ve split my integral into a sum of two integrals. The area under two curves added together is the same as adding the areas under each curve. Also, this is how you would integrate a polynomial.]
g(x)h(x) + c = ∫g’(x)h(x).dx + ∫g(x)h’(x).dx [I’ve substituted f(x) = (g(x)h(x)]
g(x)h(x) + c - ∫g’(x)h(x).dx = ∫g(x)h’(x).dx [subtract ∫g’(x)h(x).dx from both sides]
∫g(x)h’(x).dx = g(x)h(x) + c - ∫g’(x)h(x).dx [Flip it so that the equation reads in the direction we want it to. It’s ot needed; it just makes things look a little bit nicer]
∫g(x)h’(x).dx = g(x)h(x) - ∫g’(x)h(x).dx [We don’t need the “+c” because there is an integral on the right side of the equation which will add a constant anyway.]
∫g(x)h’(x).dx = g(x)h(x) - ∫g’(x)h(x).dx This is the formula for integration by parts.
