7.3: Antidifferentiation (Integration )by Parts
Proving the integration by parts formula:
Let 𝑢 and 𝑣 be differentiable functions of 𝑥. Then
𝑑
(𝑢𝑣)
𝑑𝑥
=
Thus, the integration by parts formula is
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
**We typically use integration by parts to evaluate integrals which can be written as a product of
functions that we can’t evaluate using a 𝑢-sub.
**When integration by parts works easily, we choose the function 𝑢 so that each time we differentiate
it, we eventually get zero.
Example 1: Evaluate each integral.
(a) ∫ 𝑥 cos 𝑥 𝑑𝑥
(b) ∫ 5𝑥 𝑒 2𝑥 𝑑𝑥
**Repeated integration by parts:
(c) ∫ 𝑥 2 𝑒 𝑥 𝑑𝑥
**Good news! Repeated integration by parts can be done using a table:
(d) ∫ 5𝑥 4 sin 𝑥 𝑑𝑥
𝑢 and its derivatives
𝑑𝑣 and its integrals
𝑥4
sin 𝑥
4𝑥 3
− cos 𝑥
12𝑥 2
− sin 𝑥
24𝑥
cos 𝑥
24
sin 𝑥
0
− cos 𝑥
**After forming your table, form cross products of each result and alternate the signs starting with a
positive value.
**Integration by Parts where neither function reduces to zero upon differentiation
 These problems either can’t be solved or must be solved by solving the equation for an
unknown integral. Any choice can be made for 𝒖 and 𝒅𝒗 however, do not switch your choice
later in the problem!
(e) ∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥
***Finding an antiderivative for ln 𝑥. This will prove that integration by parts can be applied to any type
of integral. (However, it may not always work!)
(f) ∫ ln 𝑥 𝑑𝑥
Add this result to your mental Calculus arsenal!!!
7.3 Assignment (Day 1): pg 350 1-10 (Hint for 9 and 10…you may need to make an atypical choice for 𝑢
and 𝑑𝑣) **Use tabular integration by parts only when more than one interation of IBPs is required