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Integration by Parts

Calculus 2 · Chapter 1: Review & Advanced Integration TechniquesSection 1.2 · lesson 2 of 5411 min · free

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In this lesson

The product rule, read backward.

  • Derive the parts formula from the product rule
  • Use LIATE to choose u and dv without guessing
  • Handle repeated applications with the tabular method
  • Solve the circular case algebraically, and apply limits correctly

Transcript

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MAT-265 - Calculus 2 · Section 1.2

Integration by Parts

A spacecraft's thrusters fire, and the onboard computer needs to know how far it drifted. The velocity, in a simplified model, is t times e to the minus t. Speed that builds early, then fades as fuel pressure drops. Distance is the integral of velocity. So somebody has to evaluate the integral of t e to the minus t. And here's the problem. Every rule you've learned so far falls flat. The power rule handles the t. The exponential rule handles the e to the minus t. There is no rule for their product. Substitution doesn't rescue you either — letting u be minus t handles the exponential, but strands that t with nowhere to go. That gap is exactly what integration by parts was built to close. It's the integral counterpart of the product rule. And alongside substitution, it's one of the two most important techniques in all of integral calculus.

The Gap

When the Power Rule Runs Out

Let's look at why nothing you already know will work here. You have a product. Two functions multiplied together. Substitution works when the integrand contains an inner function and a constant multiple of its derivative. Look at t e to the minus t — that's just not the shape. The derivative of minus t is minus one — so d u carries no t, and the stray t in the integrand has nothing to absorb it. So we need something genuinely new. And here's the good news. You already know the rule it comes from.

Derivation

It Is the Product Rule, Read Backward

Integration by parts isn't a trick somebody invented. It's the product rule, read backward. Start with the product rule. The derivative of u times v is u times v prime, plus v times u prime. Now integrate both sides. Integration undoes differentiation, so the left side collapses straight back to the product u v. On the right you're left with two integrals. Solve for the first one. Write it in differential notation, and here's the formula. The integral of u d v equals u v, minus the integral of v d u. Now — how should you read that? Read it as a trade. You're swapping one integral for another. And that trade is only worth making when the new integral is simpler than the one you started with. That is the entire skill of integration by parts. Splitting the integrand so that what's left over is easier than what you began with.

Worked Example 1

Back to the Spacecraft

So let's go back to the spacecraft and watch it work. The integral of t e to the minus t, d t. Split the integrand. Let u be t, and let d v be e to the minus t d t. Differentiate u: d u is just d t. Integrate d v: v is minus e to the minus t. Now substitute into the formula. u v gives you minus t e to the minus t. And then you subtract the integral of v d u, which is the integral of minus e to the minus t. The two minus signs make that a plus, and what's left is elementary. The answer is minus e to the minus t, times the quantity t plus one, plus C. Now notice what actually did the work here. The original integrand had a t that no rule could handle. Choosing u equals t made that t disappear the moment you differentiated it. d u is simply d t. That's the whole game. Pick the factor that gets simpler when you differentiate it.

Choosing u and dv

The LIATE Heuristic

So how do you know which factor to call u? Because the wrong choice doesn't just slow you down — it can make the integral harder than when you started. The standard guide is LIATE. It ranks function types by how good a choice they make for u. L, logarithmic. I, inverse trigonometric. A, algebraic — that's your polynomials. T, trigonometric. E, exponential. Whichever type appears earliest in that list, make it u. Whatever's left over, together with d x, becomes d v. Here's the logic behind the ordering, and it's worth understanding rather than memorizing. Logarithmic and inverse trig functions are awkward to integrate, but they simplify dramatically when you differentiate them. So they're excellent choices for u. Exponentials and trig functions differentiate and integrate about equally easily. So they're perfectly happy sitting in d v. One caution. LIATE is a heuristic, not a law. It works for the large majority of standard problems. It is not guaranteed. But it's the right first guess almost every time.

Worked Example 2

LIATE in Action

Let's put LIATE to work. The integral of x times natural log of x. The integrand mixes an algebraic factor, x, with a logarithmic factor, log x. LIATE puts L before A. So u is log x, and d v is x d x. Differentiate: d u is one over x, d x. Integrate: v is x squared over two. Apply the formula and the second integral becomes the integral of x over two — which is about as easy as it gets. You land on x squared log x over two, minus x squared over four, plus C. Now here's the question worth sitting with. Why is log x a better choice for u than for d v? Because differentiating it gives you the tidy one over x, which tends to cancel against other factors. Whereas integrating log x would itself require integration by parts. You'd be solving a harder problem inside your problem. Choose as u the function that gets simpler when differentiated. Choose as d v the function you already know how to integrate.

Repeated Application

When Once Is Not Enough - the Tabular Method

Some integrals need parts more than once. When a polynomial of degree two or higher multiplies an exponential, each application knocks the degree down by exactly one. Take the integral of x squared e to the x. One application leaves you with x e to the x — still a product. So you go again. It works. But the bookkeeping gets tedious quickly, and tedious bookkeeping is where mistakes live. So here's a faster way. The tabular method. Build two columns. The first holds u and its successive derivatives, all the way down to zero. The second holds d v and its successive antiderivatives. For x squared e to the x, the derivative column runs x squared, then two x, then two, then zero. The antiderivative column is e to the x, every time. Now read along the diagonals. Multiply each derivative entry by the antiderivative entry one row down, and attach the sign at the start of that row. Plus, minus, plus. You get x squared e to the x, minus two x e to the x, plus two e to the x, plus C. Same answer as the long way, in about a third of the writing. The table works whenever the u column differentiates down to zero — which is true for any polynomial — and the d v column can be integrated over and over.

The Circular Case

When the Integral Comes Back Around

Now a different situation, and it's the one that panics students. What if neither factor ever vanishes when you differentiate it? An exponential times a sine, for instance. You apply parts. You get a new integral. You apply parts again. And the original integral comes right back at you. It looks like you've gone in a circle and wasted your time. You haven't. Look at what you actually have. Call the original integral I. After two applications you've shown that I equals e to the x sine x, minus e to the x cosine x, minus I. That is not a dead end. That is an equation, and I is the unknown. Add I to both sides. Two I equals e to the x sine x minus e to the x cosine x. Divide by two, add your constant, and you're done. One warning, and it matters. On the second application you have to keep the same kind of choice. Trig as u, exponential as d v, both times. If you switch conventions halfway through, you simply undo the first step and land back exactly where you started.

Definite Integrals

Limits Go on Every Term

Last piece. What changes when the integral has limits? The formula is the same, but the limits go on every term. The integral from a to b of u d v equals the bracket u v, evaluated from a to b, minus the integral from a to b of v d u. That bracket means u of b times v of b, minus u of a times v of a. The limits apply to the boundary term and to the remaining integral alike. Don't put them on one and forget the other. Let's do one. The integral from zero to one of x e to the two x. Let u be x and d v be e to the two x d x. Then d u is d x, and v is one half e to the two x. Evaluate the boundary term first: you get e squared over two. Then the remaining integral gives one quarter, times e squared minus one. Combine them and you get e squared plus one, all over four. Numerically, about two point zero nine seven. One habit worth building. Evaluate the boundary term as you go, rather than carrying a full antiderivative to the very end. It keeps the arithmetic tidy and it cuts your error rate.

Key Takeaways

What to Carry Forward

So what do you take out of this section? Integration by parts comes straight from the product rule. The integral of u d v equals u v, minus the integral of v d u. And it only succeeds when that new integral is simpler than the original. Always check that before you commit. LIATE tells you what to pick for u. Logarithmic, inverse trig, algebraic, trigonometric, exponential — earliest in the list wins. A polynomial times an exponential or a trig function may need parts several times. The tabular method organizes that whenever your u column differentiates down to zero. The circular case isn't a failure. It's an equation — collect the like terms and solve. Just keep your choice of u and d v consistent throughout. And for definite integrals, put the limits on both the boundary term and the remaining integral. Next up is section one point three, Trigonometric Integrals — where the right identity does the work that neither substitution nor parts can.

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FAQ

Questions students ask us

About Integration by Parts

What is the integration by parts formula and where does it come from?
The formula is \int u\,dv = uv - \int v\,du, and it comes from integrating both sides of the product rule (uv)' = uv' + vu' and solving for one of the integrals. It trades one integral for another, so it pays off only when \int v\,du is simpler than the original. For definite integrals, the limits apply to both the boundary term uv and the remaining integral.
How do you choose u and dv in integration by parts?
Use LIATE, which stands for logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. Whichever type appears earliest becomes u, and the rest, together with dx, becomes dv. For \int x\ln x\,dx, take u=\ln x and dv = x\,dx. This gives \frac{x^2\ln x}{2} - \frac{x^2}{4} + C. LIATE is a heuristic rather than a guarantee, but it is usually a strong first guess.
When can you use the tabular method for integration by parts?
The tabular method works when the u column differentiates down to zero, as any polynomial does, and the dv column can be integrated over and over. A polynomial u alone is not enough. In \int x^2e^{x^2}\,dx, for example, e^{x^2} has no elementary antiderivative. For \int x^2e^x\,dx, multiplying along the diagonals with signs +,-,+ gives x^2e^x - 2xe^x + 2e^x + C.
What do you do when integration by parts gives back the original integral?
Treat the result as an equation and solve for the original integral algebraically. For I = \int e^x\sin x\,dx, two applications give I = e^x\sin x - e^x\cos x - I. That means I = \frac{e^x\sin x - e^x\cos x}{2} + C. Keep the same kind of choice on both applications. Trig as u both times works, and so does e^x as u both times. Switching halfway simply undoes the first step.

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