6.11 Integrating Using Integration by Parts (BC ONLY)

Introduction

Note: this is a BC Calculus topic only. It isn’t on the AB exam.

Integration by parts is the reverse of the product rule for derivatives. You use it when the integrand is a product of two functions that doesn’t fit any u-substitution pattern.

The Integration by Parts Formula

\boxed{\int f(x) \cdot g'(x) \, dx = f(x) \cdot g(x) - \int f'(x) \cdot g(x) \, dx}

Alternative notation, using $u$ and $v$ :

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

Key pattern: the formula has

g

(not

g'

) on the right side, and it switches from

f

f'

inside the remaining integral.

Derivation from the Product Rule

Integration by parts comes straight from the product rule for derivatives.

Product rule:

\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)

Solve for $f(x)g'(x)$ :

f(x)g'(x) = \frac{d}{dx}[f(x) \cdot g(x)] - f'(x)g(x)

Integrate both sides:

\int f(x)g'(x) \, dx = \int \frac{d}{dx}[f(x) \cdot g(x)] \, dx - \int f'(x)g(x) \, dx

\int f(x)g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx

That’s the integration by parts formula.

Choosing $f$ and $g'$ (or $u$ and $dv$ )

Critical decision: you have to figure out which part of the integrand is

f

and which is

g'

Strategy for Choosing $f$

Choose

f

to be the function that gets simpler when you differentiate it.

Good choices for $f$ :

Polynomials like $x$ , $x^2$ , $x^4$ , which differentiate down to smaller powers

Logarithms like $\ln x$ , which differentiates to $\frac{1}{x}$

Inverse trig like $\tan^{-1}x$ , which differentiates to something algebraic

Strategy for Choosing $g'$

Choose

g'

to be the function that’s easy to integrate.

Good choices for $g'$ :

Exponential functions like $e^x$

Trig functions like $\sin x$ and $\cos x$

Simple powers

LIATE Rule (Priority Order)

When in doubt, choose

f

using this priority (first on the list is the best choice for

f

Logarithmic functions

Inverse trig functions

Algebraic functions (polynomials)

Trigonometric functions

Exponential functions

Example 1: Basic Integration by Parts

Evaluate

\int x \sin(x) \, dx

Step 1: Identify

f

and

g'

Which one gets simpler when differentiated?

The derivative of $x$ is $1$ , which is simpler

The derivative of $\sin x$ is $\cos x$ , which isn’t simpler

So choose

f(x) = x

and

g'(x) = \sin x

Step 2: Find

f'

and

g

f(x) = x \quad \Rightarrow \quad f'(x) = 1

g'(x) = \sin x \quad \Rightarrow \quad g(x) = -\cos x

Step 3: Apply the formula.

\int f \cdot g' \, dx = f \cdot g - \int f' \cdot g \, dx

\int x \sin(x) \, dx = x \cdot (-\cos x) - \int 1 \cdot (-\cos x) \, dx

= -x\cos x - \int (-\cos x) \, dx = -x\cos x + \int \cos x \, dx

Step 4: Integrate what’s left.

= -x\cos x + \sin x + C

Verification tip: take the derivative of your answer to check.

\frac{d}{dx}[-x\cos x + \sin x] = -\cos x + x\sin x + \cos x = x\sin x \quad \checkmark

Example 2: A Definite Integral

Evaluate

\int_1^e x\ln x \, dx

For a definite integral, evaluate the

f \cdot g

part at the limits too:

\boxed{\int_a^b f(x) \cdot g'(x) \, dx = \Big[f(x) \cdot g(x)\Big]_a^b - \int_a^b f'(x) \cdot g(x) \, dx}

Step 1: Identify

f

and

g'

. By LIATE, the logarithm comes first, so

f(x) = \ln x

and

g'(x) = x

f(x) = \ln x \quad \Rightarrow \quad f'(x) = \frac{1}{x}

g'(x) = x \quad \Rightarrow \quad g(x) = \frac{x^2}{2}

Step 2: Apply the formula.

\int_1^e x\ln x \, dx = \left[\frac{x^2}{2}\ln x\right]_1^e - \int_1^e \frac{1}{x} \cdot \frac{x^2}{2} \, dx = \left[\frac{x^2}{2}\ln x\right]_1^e - \int_1^e \frac{x}{2} \, dx

Step 3: Evaluate. Remember

\ln e = 1

and

\ln 1 = 0

= \left(\frac{e^2}{2} \cdot 1 - \frac{1}{2} \cdot 0\right) - \left[\frac{x^2}{4}\right]_1^e = \frac{e^2}{2} - \left(\frac{e^2}{4} - \frac{1}{4}\right) = \frac{e^2 + 1}{4} \approx 2.097

Why not

f = x

? Then

g' = \ln x

, and you’d have to integrate

\ln x

before you could even start, which takes integration by parts all over again.

Tabular Integration (Repeated Integration by Parts)

When one function has to be differentiated several times before it hits zero, tabular integration is a shortcut.

When to Use It

Use it when:

One part is a polynomial (so it can be differentiated down to zero)

The other part is easy to integrate over and over, like $\sin x$ , $\cos x$ , or $e^x$

Example: $\int x^4 \sin(x) \, dx$

Step 1: Set up the table.

Column 1 (differentiate $f$ ): keep differentiating until you reach 0.

\begin{array}{c} x^4 \\ 4x^3 \\ 12x^2 \\ 24x \\ 24 \\ 0 \end{array}

Column 2 (integrate $g'$ ): integrate the same number of times.

\begin{array}{c} \sin x \\ -\cos x \\ -\sin x \\ \cos x \\ \sin x \\ -\cos x \end{array}

Step 2: Draw diagonal arrows from each term in Column 1 to the term one row below in Column 2.

$x^4 \rightarrow -\cos x$

$4x^3 \rightarrow -\sin x$

$12x^2 \rightarrow \cos x$

$24x \rightarrow \sin x$

$24 \rightarrow -\cos x$

Step 3: Assign alternating signs, starting with

(+)

\begin{array}{cc} (+) & x^4 \rightarrow -\cos x \\ (-) & 4x^3 \rightarrow -\sin x \\ (+) & 12x^2 \rightarrow \cos x \\ (-) & 24x \rightarrow \sin x \\ (+) & 24 \rightarrow -\cos x \end{array}

Step 4: Multiply along the arrows and combine with the signs.

\int x^4\sin x \, dx = (+)(x^4)(-\cos x) + (-)(4x^3)(-\sin x) + (+)(12x^2)(\cos x)

+ (-)(24x)(\sin x) + (+)(24)(-\cos x) + C

= -x^4\cos x + 4x^3\sin x + 12x^2\cos x - 24x\sin x - 24\cos x + C

Why this works: it’s exactly what you’d get from doing integration by parts four times in a row, just organized efficiently.

Note on Multiple Choice

On AP exams, the answer choices sometimes include an intermediate step with a remaining integral in it:

-x^4\cos x + 4x^3\sin x + \int (\ldots) \, dx

If you spot that pattern in the choices, don’t automatically finish every step. You might save yourself some time.

Key Takeaways

Formula: $\int f \cdot g' \, dx = f \cdot g - \int f' \cdot g \, dx$

Choose $f$ to be the function that gets simpler when differentiated (polynomials, $\ln x$ , inverse trig)

Choose $g'$ to be the function that’s easy to integrate ( $e^x$ , $\sin x$ , $\cos x$ )

LIATE rule: logarithmic, then inverse trig, then algebraic, then trig, then exponential

Don’t just go left to right. Pick $f$ and $g'$ strategically

It comes from the product rule. Integration by parts reverses it

You may need multiple applications. Sometimes the leftover integral needs parts again

Tabular integration is an efficient shortcut when one part differentiates down to zero

Tabular signs alternate: $(+), (-), (+), (-), \ldots$ starting with $(+)$

Check your work by differentiating your answer to see if you get the original integrand

BC only. Not tested on the AB exam

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