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Integration by Parts, Including Repeated Application

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

Integration by Parts

Including Repeated Application

Trading one integral for a simpler one — the product rule read backwards

1. Derivation — The Product Rule Read Backwards

Starting Point: The Product Rule

Recall the product rule for derivatives. If $f(x)$ and $g(x)$ are differentiable functions, then:

$$\frac{d}{dx}\bigl[f(x)\,g(x)\bigr] = f'(x)\,g(x) + f(x)\,g'(x)$$
Step 1: Integrate Both Sides

Integrate every term with respect to $x$. The left side "undoes" the derivative:

$$f(x)\,g(x) = \int f'(x)\,g(x)\,dx + \int f(x)\,g'(x)\,dx$$
Step 2: Rearrange

Solve for the second integral:

$$\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int f'(x)\,g(x)\,dx$$
Two Notational Forms
Differential Notation ($u$-$v$ form)

Let $u = f(x)$, $dv = g'(x)\,dx$

Then $du = f'(x)\,dx$, $v = g(x)$

$$\boxed{\int u\,dv = u\,v - \int v\,du}$$

This is the form you will use most often.

Explicit Derivative Notation
$$\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int f'(x)\,g(x)\,dx$$

Useful when you want to see the functions and their derivatives explicitly.

What Does the Formula Actually Do?

Integration by parts trades one integral for another, plus a boundary product $uv$. The hope is that the new integral $\int v\,du$ is simpler than the original $\int u\,dv$. If it isn't, you chose the wrong parts!

2. Choosing the Parts — The LIATE Guideline

The strategic question is: which factor should be $u$ (differentiated) and which should be $dv$ (antidifferentiated)?

The LIATE Priority

A useful heuristic — not an unbreakable law — ranks function types by how much they want to be $u$ (differentiated):

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