Integration by parts

Integration by parts is a method for integrating a product of two functions, such as x eˣ or x sin x. The formula is ∫u dv = uv − ∫v du: you choose one factor to differentiate (u) and one to integrate (dv), and the new integral on the right should be easier than the one you started with. It is the product rule for derivatives, run backward.

Integration by parts formula

UK textbooks and the Edexcel A-level formula booklet write the same thing with derivatives:

Where the formula comes from

Start with the product rule for two functions u and v:

Integrate both sides with respect to x. The left side just gives back uv:

Rearrange, and you have the formula. So every integration by parts problem is a product rule problem in reverse.

How to do integration by parts

1. Choose u and dv. Pick u so that it gets simpler when you differentiate it. The rest of the integrand, including dx, is dv.
2. Find du and v. Differentiate u and integrate dv.
3. Substitute into the formula: uv − ∫v du.
4. Do the new integral. If it is still a product, you may need integration by parts again.
5. Add + C at the end of an indefinite integral.

Choosing u with LIATE

Choosing well is most of the skill. LIATE is a rule of thumb: pick u from whichever type of function comes first in this list.

Functions near the top get simpler when differentiated: ln x becomes 1/x, and x² becomes 2x and then 2. Functions near the bottom are easy to integrate and don't get worse: eˣ stays eˣ. LIATE is a guide, not a law, but it works for nearly every textbook problem.

Integration by parts examples

Example 1: ∫x eˣ dx

Algebraic comes before exponential, so u = x and dv = eˣ dx. Then du = dx and v = eˣ.

Check by differentiating: the product rule gives eˣ + x eˣ − eˣ = x eˣ. It works.

Example 2: ∫x cos x dx

u = x and dv = cos x dx, so du = dx and v = sin x.

Example 3: ∫arcsin x dx

There seems to be only one function here, but you can write the integrand as arcsin x times 1. Take u = arcsin x and dv = dx, so du = dx/√(1 − x²) and v = x.

The new integral is a quick substitution with w = 1 − x², and it comes to −√(1 − x²). So:

The same "times 1" trick works for the integral of ln x: with u = ln x and dv = dx, it comes out as x ln x − x + C.

Example 4: ∫(2x + 1) ln x dx

Logarithmic beats algebraic, so u = ln x and dv = (2x + 1) dx. Then du = (1/x) dx and v = x² + x.

Divide before you integrate: (x² + x)/x is just x + 1.

If you had chosen u = 2x + 1 instead, you would need to integrate ln x for v and the problem would get harder, not easier.

Example 5: ∫x² eˣ dx, by parts twice

u = x², dv = eˣ dx, so du = 2x dx and v = eˣ:

The new integral is example 1 times 2, so use parts again:

Each round lowers the power of x by one, so a polynomial of degree n needs n rounds.

Practice choosing u before the harder cases.

Integration by parts with a loop: ∫eˣ sin x dx

Some integrals never get simpler, but they come back to where they started. Then you solve for the integral like an unknown. Call it I.

Round 1: u = sin x and dv = eˣ dx, so du = cos x dx and v = eˣ:

Round 2: u = cos x and dv = eˣ dx, so du = −sin x dx and v = eˣ:

The original integral has appeared again. Add I to both sides and divide by 2:

Keep the same type of function as u in both rounds (trig here). If you switch, the second round simply undoes the first.

The tabular method

When u is a polynomial that eventually differentiates to 0, the tabular (or DI) method saves a lot of writing. List u and its derivatives in one column, dv and its integrals in the other, then multiply diagonally with alternating signs, starting with +.

For ∫x³ eˣ dx:

Multiply each entry in the middle column by the entry one row down in the right column, using the sign on its row:

Integration by parts for definite integrals

For a definite integral, apply the limits to the uv part as well:

Example 6: ∫ from 0 to π of x sin x dx

u = x and dv = sin x dx, so du = dx and v = −cos x:

Here x is the input in radians and y = x sin x. The curve stays above the x-axis between 0 and π, and the area under it is exactly π, about 3.14. You can shade this area and check the value in the Integral Calculator.

When to use integration by parts

• Use it for a product of two different types of function where one simplifies when differentiated: x eˣ, x² sin x, x ln x.
• Use it for single functions with no simple antiderivative of their own: ln x, arcsin x, arctan x.
• Try substitution first when one factor is the derivative of something inside the other, like 2x eˣ² or x cos(x²). For ∫2x eˣ² dx, the substitution w = x² is much quicker than parts.

Common mistakes

• Choosing u the wrong way round. If the new integral looks harder, swap u and dv. LIATE helps.
• Sign errors, especially with sin and cos. The integral of sin x is −cos x.
• Forgetting the limits on uv in a definite integral. [uv] from a to b must be evaluated too.
• Dropping the dx from dv. dv includes the dx, which keeps track of what you integrate.
• Forgetting + C on indefinite integrals, or adding it to v midway, which is unnecessary.
• Switching function types between rounds in a loop integral, which just undoes the first round.

Integration by parts in exams

In the US, integration by parts is in AP Calculus BC (topic 6.11) but not AP Calculus AB, and it is a core technique in college Calculus II. In the UK it is part of A-level Maths, and the Edexcel formula booklet gives the formula, so you need to know how to use it rather than memorize it.

The daily AP Calculus BC and College Calculus II & III papers are short timed practice tests of 8 questions, with a new one every day. In the UK, the same practice is A-level Maths: Calculus. If you are new to integration, start with the Calculus course.

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