Integration by Parts: When to Use It and How

You already know how to integrate $x$, and you already know how to integrate $e^x$. So why does $\int x e^x\, dx$ feel impossible?

The problem is that integration doesn't distribute over multiplication. $\int f \cdot g\, dx$ is not $\int f\, dx \cdot \int g\, dx$. You can't just integrate each piece and multiply them together. You need a different tool entirely, and that tool is integration by parts.

This post walks through where the formula comes from, how to actually choose which part is "u," and what to do in the two situations that trip people up: when you have to repeat the process, and when the process refuses to end on its own.

Multiplying two functions breaks all your usual tricks

Every integration rule you've learned so far works on a single function. The power rule handles $x^n$. The exponential rule handles $e^x$. The trig rules handle $\sin x$ and $\cos x$.

None of them say anything about what happens when two of these get multiplied together. $x e^x$ isn't $x^n$, and it isn't $e^x$ either. It's a new object, and none of your existing rules apply to it directly.

We need a rule built specifically for products. And it turns out we already have one. We've just been using it in the wrong direction.

The product rule, run backwards

Here's the move. You already know the product rule for derivatives. We write it as: if $y = uv$, then

Rearrange that one line and something useful falls out:

Now integrate both sides with respect to $x$. The integral of a derivative just undoes the derivative, so $\int (uv)'\, dx = uv$. That gives us:

Switch to the $du$/$dv$ notation you'll actually see in textbooks, where $du = u'\,dx$ and $dv = v'\,dx$, and this becomes the formula everyone calls integration by parts:

That's the whole trick. You're not learning a new operation. You're running the product rule in reverse.

Watch the formula prove itself

Algebra convinces you the formula is true. But there's a picture that makes it obvious, and it doesn't use a single symbol from calculus.

Think of $u$ and $v$ as the two side lengths of a rectangle, both growing together as some parameter increases. The rectangle's area is $u \cdot v$. As it grows, that area gets added to in exactly two ways: a strip appears where the width grew, and a strip appears where the height grew.

Drag the point along the curve. At every position, the full rectangle $u \cdot v$ is exactly covered by two puzzle pieces: the region under the curve, which is $\int v\, du$, and the region to the left of the curve, which is $\int u\, dv$. They never overlap and they never leave a gap. Add them and you always get $u \cdot v$ back, exactly.

That's the entire formula, sitting there as a picture. Rearrange "the two pieces add up to the rectangle" and you get $\int u\, dv = uv - \int v\, du$.

So which part is u?

Here's where people get stuck. The formula needs you to split your integrand into two pieces, $u$ and $dv$. Pick well and the new integral is easier than the one you started with. Pick badly and it gets worse.

A rough guide that works most of the time is called LIATE. When you're deciding what to call $u$, prefer, in this order: Logarithmic functions, Inverse trig functions, Algebraic functions like $x$ or $x^2$, Trig functions, Exponential functions. Whichever type appears first in that list becomes your $u$. Everything else becomes $dv$.

The reasoning behind LIATE is simple once you see it: you want $u$ to be a function that gets simpler when you differentiate it, and you want $dv$ to be a function that stays manageable when you integrate it. Logs and inverse trig get messier if you try to integrate them, so you never want them stuck inside an integral. Exponentials barely change no matter what you do to them, so they're happy sitting as $dv$.

Try it on $\int x e^x\, dx$. Algebraic beats exponential in LIATE, so $u = x$.

Pick the LIATE choice and $du = dx$, $v = e^x$, and the new integral is just $\int e^x\, dx$. Done in one more step. Pick it backwards and $v = x^2/2$, and the new integral has a higher power of $x$ than you started with. Keep applying the same backwards choice and the power climbs forever. It never gets easier. It only gets worse.

What if one pass through the formula isn't enough?

LIATE tells you what to do once. But some integrals need the formula applied several times before they resolve. $\int x^2 e^x\, dx$ is a good example: after one pass you're left with $\int 2x\, e^x\, dx$, still a product, still not done.

You could keep applying the full formula by hand at every step, tracking $u$, $du$, $v$, and $dv$ each time. Or you could use a shortcut that does the bookkeeping for you: the tabular method, sometimes called the DI method.

Put the part you'd normally call $u$ in a "D" column and keep differentiating it until you hit zero. Put the part you'd normally call $dv$ in an "I" column and keep integrating it, matching the D column row for row. Then multiply diagonally, one row of D against the next row of I, alternating plus and minus signs starting with plus.

Step through the table. $x^2$ differentiates down to $0$ in three steps, so the table has exactly three usable rows, and each one hands you a term of the answer. Add them up and you get $x^2 e^x - 2xe^x + 2e^x + C$, the same answer you'd reach by applying the formula three separate times, just without re-deriving $u$ and $v$ at every step.

The tabular method only works this cleanly when one column eventually hits zero, which happens whenever the "D" side is a polynomial. Polynomials always differentiate down to nothing eventually. Not every product behaves this way.

What if the integral never gets simpler?

Try $\int e^x \sin x\, dx$. Neither $e^x$ nor $\sin x$ ever disappears when you differentiate or integrate them. There's no polynomial here to grind down to zero. The tabular method's trick doesn't apply.

Apply integration by parts anyway and watch what happens.

Step through it. The first pass turns your integral into $e^x \sin x$ minus a new integral, $\int e^x \cos x\, dx$. Fine, apply the formula again to that one. But the second pass hands you back $\int e^x \sin x\, dx$, the exact integral you started with.

Call your original integral $I$. After two passes, the algebra reads $I = e^x \sin x - e^x \cos x - I$. You now have an equation for $I$, with $I$ sitting on both sides. Add $I$ to both sides, divide by 2, and you're done:

No further integration needed. You solved it the way you'd solve $2I = 6$ for $I$, by treating the unknown integral as a variable and doing algebra on it. Whenever a repeated application of integration by parts hands you back your original integral, this is the way out.

A quick way to know when to reach for it

Reach for integration by parts when you're looking at a product of two functions from different families: an algebraic function times an exponential, a function times a log, or an exponential times a trig function, and none of your other techniques apply. Use LIATE to pick $u$. If the new integral is still a product, either repeat the process (the tabular method helps when a polynomial is involved) or watch for a loop back to the original integral, and solve algebraically if you find one.

The short version

Integration by parts is the product rule for derivatives, read backwards. Split your product into $u$ and $dv$, and you trade the integral you can't do for a different one, hopefully easier, that you can.

LIATE tells you which part to call $u$: pick the one that gets simpler when you differentiate it. Sometimes one pass is enough. Sometimes you need several, and the tabular method tracks the bookkeeping for you. And sometimes the integral doesn't shrink at all. It loops back to where it started, and the way out is algebra, not more calculus.

Three integrals, three different endings, one formula behind all of them.