When should I use integration by parts instead of substitution?

Use substitution when the integrand contains a function and (a multiple of) its derivative, so a single u collapses the whole expression. Use integration by parts when you have a product of two unlike functions — such as a polynomial times an exponential, or a logarithm times anything — where no substitution simplifies things. If substitution fails and you see a product, try parts.

What does LIATE stand for and is it always correct?

LIATE stands for Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — the priority order for choosing u. It works for nearly every AP-level problem, but it is a heuristic, not a theorem. The real test is whether your choice makes v\,du simpler than the original. If it does not, switch which factor you call u.

How do I integrate ln(x)?

Treat x as a product with dv = dx. Set u = x and dv = dx, so du = 1/xdx and v = x. Then x\,dx = x x - x·1/xdx = x x - x + C. This is a standard result worth memorizing.

Why does the integral sometimes reappear on the right side?

For integrands like e^x x or e^x x, neither factor ever differentiates to zero, so repeated parts cycles back to the original integral. When that happens, set the integral equal to a variable like I, collect the I terms on one side, and solve algebraically. This is the intended method, not a mistake.

AP-CALCBC-6.11

U6.11 Integration by Parts

Master integration by parts for AP Calculus BC: learn the formula ∫u dv = uv − ∫v du, the LIATE rule for picking u, tabular method, and definite integrals.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.11 Integration by Parts, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U6.11 Integration by Parts free

What this lesson covers

Substitution reverses the chain rule, but what undoes the product rule? That is exactly the job of integration by parts, one of the most tested antidifferentiation techniques on the BC exam. Whenever you see a product of two unlike functions — say

x e^x

x^2 \ln x

, or

e^x \cos x

— substitution stalls and integration by parts takes over.

In this lesson you will derive the formula from the product rule, learn the LIATE heuristic for choosing

u

wisely, practice the tabular shortcut for repeated applications, and see how the technique appears in both indefinite and definite integrals. By the end you will be able to recognize which integrals call for parts and execute them cleanly under exam time pressure.

The Formula and Where It Comes From

Integration by parts is the product rule run in reverse. Start with the product rule for two differentiable functions

u

and

v

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.

Integrate both sides with respect to

x

uv = \int u\,dv + \int v\,du.

Rearranging gives the working formula you must memorize:

\int u\,dv = uv - \int v\,du.

The strategy is to split the integrand into two pieces: a part you call

u

(which you will differentiate) and a part you call

dv

(which you will integrate). You then compute

du

by differentiating and

v

by integrating. The goal is to trade the original integral

\int u\,dv

for a new, simpler integral

\int v\,du

.

The entire art of the method lies in choosing

u

and

dv

so that the new integral is easier than the one you started with. A poor choice can make the integral harder or send you in circles. That is why a systematic guideline — LIATE — is so valuable, and why you should always check that

\int v\,du

is genuinely simpler before committing to your split.

LIATE: Choosing u Strategically

LIATE is a mnemonic that ranks function types by how good a choice they are for

u

. The function type appearing earliest in the list should be chosen as

u

; whatever remains becomes

dv

Letter	Function type	Example
L	Logarithmic	$\ln x$
I	Inverse trig	$\arctan x$
A	Algebraic (polynomials)	$x^2$ , $3x$
T	Trigonometric	$\sin x$ , $\cos x$
E	Exponential	$e^x$

The reasoning: functions near the top (logs, inverse trig) are hard to integrate but easy to differentiate, so they make excellent choices for

u

. Functions near the bottom (exponentials, trig) integrate cleanly, so they make good

dv

choices.

For

\int x\ln x\,dx

, the logarithm outranks the algebraic term, so let

u = \ln x

and

dv = x\,dx

. For

\int x e^x\,dx

, algebraic beats exponential, so

u = x

and

dv = e^x\,dx

; differentiating

x

gives the constant

1

, collapsing the remaining integral.

A common misconception is that LIATE is an unbreakable law. It is a heuristic that works the vast majority of the time on exam problems, but the true test is whether your choice makes

\int v\,du

simpler. If it does not, swap your choices.

Repeated Parts and the Tabular Method

Some integrals require applying integration by parts more than once. When

u

is a polynomial of degree

n

, you may need

n

applications until the polynomial differentiates down to a constant. The tabular method organizes this efficiently.

Consider

\int x^2 e^x\,dx

. Build two columns: repeatedly differentiate

u = x^2

until you reach

0

, and repeatedly integrate

dv = e^x\,dx

Sign	Derivatives of $x^2$	Integrals of $e^x$
$+$	$x^2$	$e^x$
$-$	$2x$	$e^x$
$+$	$2$	$e^x$
$-$	$0$	$e^x$

Multiply diagonally, alternating signs starting with plus:

\int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2e^x + C.

A second special case is the cyclic integral, such as

\int e^x\sin x\,dx

. Here neither factor ever becomes a constant. After applying parts twice, the original integral reappears on the right side. Treat it as an unknown, add it to both sides, and solve algebraically. This algebraic trick is a favorite on free-response questions because it tests whether you recognize the pattern rather than integrating forever.

Definite Integrals and Exam Pitfalls

For a definite integral, apply the formula with limits attached to every term:

\int_a^b u\,dv = \left[uv\right]_a^b - \int_a^b v\,du.

Evaluate the boundary term

uv

at the limits immediately, then handle the remaining integral. A frequent error is forgetting to evaluate the

uv

term at both bounds, or applying the limits only at the very end.

Other pitfalls to watch: dropping the constant of integration on indefinite problems; sign errors from the minus sign in front of

\int v\,du

(this is the single most common mistake); and choosing

v

incorrectly by adding an unnecessary constant when you integrate

dv

(always take the simplest antiderivative).

The AP exam tests this topic in several ways. Multiple-choice items may ask you to identify the correct choice of

u

and

dv

, or to complete a single application of the formula. Free-response questions often embed integration by parts inside a larger problem — for instance, finding the area under

y = x\ln x

or computing an accumulation function. The integral

\int \ln x\,dx = x\ln x - x + C

deserves special memorization: choose

u = \ln x

and

dv = dx

, a case where

dv

is just

dx

and

v = x

. This appears frequently and catches students who forget that

\ln x

has no elementary antiderivative without parts.

Key terms

Integration by parts.: A technique that reverses the product rule, given by $\int u\,dv = uv - \int v\,du$ , used to integrate products of unlike functions.
LIATE.: A mnemonic (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) ranking which factor to choose as $u$ ; earlier in the list means better choice for $u$ .
u and dv.: The two parts an integrand is split into: $u$ is differentiated to get $du$ , and $dv$ is integrated to get $v$ .
Tabular method.: A shortcut for repeated integration by parts, using a table of successive derivatives and integrals with alternating signs, ideal when $u$ is a polynomial.
Cyclic integral.: An integral (like $\int e^x\sin x\,dx$ ) where the original integral reappears after repeated parts, solved by treating it as an unknown and using algebra.
Boundary term.: The $uv$ portion of a definite integration by parts, evaluated as $[uv]_a^b$ at the limits of integration.

Worked example

Evaluate

\int x\cos x\,dx

First identify the two factors: an algebraic term

x

and a trigonometric term

\cos x

. By LIATE, Algebraic (A) outranks Trigonometric (T), so let

u = x

and

dv = \cos x\,dx

.

Next compute the pieces. Differentiate

u

du = dx

. Integrate

dv

v = \sin x

(take the simplest antiderivative, no added constant).

Apply the formula

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

\int x\cos x\,dx = x\sin x - \int \sin x\,dx.

The new integral is simpler than the original — that confirms the LIATE choice was correct. Now integrate

\int \sin x\,dx = -\cos x

\int x\cos x\,dx = x\sin x - (-\cos x) + C = x\sin x + \cos x + C.

To check, differentiate the answer:

\frac{d}{dx}(x\sin x + \cos x) = \sin x + x\cos x - \sin x = x\cos x

, which matches the integrand. The answer is

x\sin x + \cos x + C

Practice questions

For the integral

\int x^3\ln x\,dx

, which choice of

u

and

dv

follows the LIATE heuristic?

$u = x^3$ , $dv = \ln x\,dx$
$u = \ln x$ , $dv = x^3\,dx$
$u = x^3\ln x$ , $dv = dx$
$u = \ln x\,dx$ , $dv = x^3$

Answer: $u = \ln x$ , $dv = x^3\,dx$

LIATE ranks Logarithmic above Algebraic, so the

\ln x

factor should be

u

. This is also practical:

\ln x

is hard to integrate but easy to differentiate (

du = \frac{1}{x}dx

), while

x^3

integrates cleanly to

v = \frac{x^4}{4}

. The resulting integral

\int \frac{x^4}{4}\cdot\frac{1}{x}dx = \int\frac{x^3}{4}dx

is straightforward.

Evaluate the definite integral

\int_1^e \ln x\,dx

. Show your reasoning.

Answer: $1$

Use integration by parts with

u = \ln x

and

dv = dx

, so

du = \frac{1}{x}dx

and

v = x

. Then

\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}dx = x\ln x - x

. Evaluate from

1

e

: at

x = e

e\ln e - e = e\cdot 1 - e = 0

; at

x = 1

1\cdot\ln 1 - 1 = 0 - 1 = -1

. Subtract:

0 - (-1) = 1

Find

\int e^x\cos x\,dx

Answer: $\frac{e^x(\sin x + \cos x)}{2} + C$

Let

u = \cos x

dv = e^x dx

, giving

\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx

. Apply parts again on the new integral with

u = \sin x

dv = e^x dx

\int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx

. Substituting back, let

I = \int e^x\cos x\,dx

I = e^x\cos x + e^x\sin x - I

. So

2I = e^x(\cos x + \sin x)

, giving

I = \frac{e^x(\sin x + \cos x)}{2} + C

FAQ

When should I use integration by parts instead of substitution?: Use substitution when the integrand contains a function and (a multiple of) its derivative, so a single $u$ collapses the whole expression. Use integration by parts when you have a product of two unlike functions — such as a polynomial times an exponential, or a logarithm times anything — where no substitution simplifies things. If substitution fails and you see a product, try parts.
What does LIATE stand for and is it always correct?: LIATE stands for Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — the priority order for choosing $u$ . It works for nearly every AP-level problem, but it is a heuristic, not a theorem. The real test is whether your choice makes $\int v\,du$ simpler than the original. If it does not, switch which factor you call $u$ .
How do I integrate ln(x)?: Treat $\ln x$ as a product with $dv = dx$ . Set $u = \ln x$ and $dv = dx$ , so $du = \frac{1}{x}dx$ and $v = x$ . Then $\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}dx = x\ln x - x + C$ . This is a standard result worth memorizing.
Why does the integral sometimes reappear on the right side?: For integrands like $e^x\sin x$ or $e^x\cos x$ , neither factor ever differentiates to zero, so repeated parts cycles back to the original integral. When that happens, set the integral equal to a variable like $I$ , collect the $I$ terms on one side, and solve algebraically. This is the intended method, not a mistake.

Learn this with a teacher, not a page

The Crimsora tutor teaches U6.11 Integration by Parts live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free