Integration by Parts Step by Step: How to Solve It Correctly

Integration by parts is a technique for integrating products of functions, and it works by trading one integral for a simpler one. The method comes directly from the product rule for derivatives and is usually written as ∫u dv = uv − ∫v du. To apply it, you pick which part of the integrand to call u and which to call dv, differentiate u, integrate dv, then plug everything into the formula. The whole process becomes straightforward once you know how to choose u and dv wisely — that choice is what makes integration by parts step by step reliable rather than guesswork.

Why the Formula Works

The rule isn’t arbitrary. It falls straight out of the product rule, which says the derivative of a product uv is u′v + uv′. Rearranging and integrating both sides gives the integration by parts formula. Khan Academy covers the derivation in detail if you want to see each algebraic move.

Because it’s derived from the product rule, the technique is exact, not an approximation. You’re rewriting one integral in terms of another that’s typically easier to evaluate. When the new integral is genuinely simpler, you win. When it isn’t, the method can loop back on itself, which is a sign you picked u and dv poorly.

The formula has one job: move the derivative off one factor and onto the other. That’s the entire strategy in a sentence.

Choosing u and dv Correctly

The most common mistake is choosing u and dv in a way that makes the second integral harder than the first. A simple priority order, often called LIATE, helps you decide. Pick u from whichever category comes first in this list:

• L — Logarithmic functions (like ln x)
• A — Algebraic functions (like x, x²)
• T — Trigonometric functions (like sin x, cos x)
• E — Exponential functions (like eˣ)

Whichever category appears earlier, that’s your u. Everything else becomes dv. The reasoning: logs and algebraic terms simplify when you differentiate them, while exponentials and trig functions stay manageable when you integrate them.

A Worked Example

Take ∫x·eˣ dx. Using LIATE, algebraic (x) beats exponential (eˣ), so u = x and dv = eˣ dx. Differentiating gives du = dx, and integrating gives v = eˣ. Plugging into the formula:

∫x·eˣ dx = x·eˣ − ∫eˣ dx = x·eˣ − eˣ + C

The remaining integral was trivial, which confirms the choice was good. If you had instead set u = eˣ and dv = x dx, you’d end up with an integral involving x² that’s messier — a clear downgrade. This kind of check, asking whether the new integral is simpler, is the single best habit to develop. MIT OpenCourseWare has worked examples that reinforce how much the u/dv choice matters.

Common Mistakes to Avoid

Most errors come from a few predictable slips. Watching for them saves a lot of frustration.

Forgetting the sign on ∫v du. The formula has a minus sign, and it’s easy to drop. Note also that v absorbs no constant from the integration step — you can add the + C at the very end.

Choosing u and dv backwards. If your new integral is worse than the original, stop and swap. The x power should be going down, not up.

Not differentiating fully. When u = x², then du = 2x dx, not 2x. Keep the dx; it matters when the next integral is set up.

Stopping too early. Some integrals, like ∫x²·sin x dx, need the method twice. If a product remains, you’re probably not done.

A quick self-check after each step: is the integral I’m facing now easier than the one I started with? If yes, continue. If no, reconsider your choice of u before grinding through more algebra.

Handling Definite Integrals

Everything so far assumed indefinite integrals, where the constant of integration matters. For a definite integral, the structure changes slightly but the strategy doesn’t. You evaluate the boundary term at both limits before tackling the remaining integral:

∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du

The square brackets mean substitute the upper limit, then subtract the result at the lower limit. Work the boundary term out completely first, then move to the second integral — mixing the two in one line invites arithmetic errors.

A useful shortcut appears when the boundary term vanishes. If uv equals zero at both limits, the entire first term disappears and you’re left with just −∫v du. This happens often with integrals over symmetric intervals or when u contains a factor that’s zero at an endpoint, and it can cut the work in half.

When to Use Another Method

Integration by parts is powerful, but it isn’t always the right tool. Substitution often handles products where one factor is the derivative of another’s inside function — say, ∫2x·cos(x²) dx. Here the 2x is the derivative of x², so a u-substitution collapses the integral instantly. Forcing the parts method onto it would work but take longer.

One case surprises many students. An integral like ∫ln x dx looks like it has no product at all, but setting dv = dx and u = ln x makes the formula work perfectly, yielding x·ln x − x + C. Neither substitution nor a simpler trick applies; parts is the right call.

The genuine skill here isn’t the arithmetic — it’s the judgment call at the start about which function to pick as u. Before writing anything down, decide which factor you want to differentiate and which you want to integrate. Then follow a fixed sequence: identify u and dv, compute du and v, substitute into ∫u dv = uv − ∫v du, and evaluate what’s left. If a product remains, repeat the sequence; if the original integral reappears, solve for it algebraically. With a handful of practice problems, the pattern recognition becomes automatic.

Key Takeaways

• Integration by parts reverses the product rule and is written ∫u dv = uv − ∫v du.
• Use the LIATE priority order to choose u: logarithmic, algebraic, trigonometric, then exponential.
• The goal is always a simpler second integral; if it gets harder, swap your u and dv.
• Some problems need two or more passes, and the x power should drop each time.
• When the original integral reappears, solve for it algebraically.
• Drop the minus sign at your peril, and add the + C only at the very end.