How do you derive the formula for integration by parts?

Question

Aurora Alvarado · Accepted Answer

The integration by parts formula is derived directly from the product rule for differentiability.

If #f# and #g# are continuously differentiable everywhere, then we can differentiate their product (using the product rule):

# d/dx (fg) = (f)( d/dx g) + (d/dx f)( g) #

# :. d/dx (fg) = f \ (dg)/dx + g \ (df)/dx #

# :. f \ (dg)/dx = d/dx (fg) - g \ (df)/dx #

Now simply integrate wrt #x#:

# int \ f \ (dg)/dx \ dx = int \ d/dx (fg) \ dx - int \ g \ (df)/dx \ dx #

From which we get the IBP formula:

# int \ f \ (dg)/dx \ dx = fg - int \ g \ (df)/dx \ dx #

Emilia Aguilar · Answer

undefined To derive the formula for integration by parts, we start with the product rule for differentiation:

$$
(uv)' = u'v + uv'
$$

Then, we integrate both sides with respect to x:

$$
\int (u v)' dx = \int (u'v + uv') dx
$$

Using the fundamental theorem of calculus, the left side becomes just $uv$, yielding:

$$
uv = \int u'v dx + \int uv' dx
$$

Rearranging this equation, we get the integration by parts formula:

$$
\int u dv = uv - \int v du
$$

This formula allows us to integrate a product of two functions by selecting one function to differentiate and the other to integrate, facilitating the integration process.