🎓 University Year 1 · Lesson 3 of 15
Integration by Parts: Running the Product Rule Backwards
When an integrand is a product of two different kinds of function, trade the integral you cannot do for one you can.
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The big idea: The product rule tells you how to differentiate a product. Integration by parts is that rule read in reverse: it lets you integrate a product by shifting the derivative from one factor onto the other. You pick one factor to differentiate (u) and the other to integrate (dv), and if you choose well the new integral ∫v du is simpler than the one you started with.
Some products resist substitution
Substitution handles integrals where one factor is (almost) the derivative of the other. But what about ∫ x ex dx? Here x and ex are unrelated — neither is the derivative of the other — and substitution stalls. We need a different lever.
That lever is the product rule, run backwards. Differentiating a product spreads a derivative across two factors; integrating by parts lets us move that derivative from the factor we cannot integrate to the one we can.
From product rule to by-parts
Start with the product rule for two functions u and v: (uv)′ = u′v + uv′. Integrate both sides. The left side integrates back to uv. Rearranging to isolate one of the two integrals on the right gives the by-parts formula.
\[ \int u\,dv = uv-\int v\,du \]
Integration by parts: trade ∫u dv for uv minus ∫v du.
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You choose who gets differentiated
Split the integrand into u (which you will differentiate, via du) and dv (which you will integrate, via v). The whole art is the split: a good choice makes ∫v du easier than the original; a bad choice makes it harder.
🎮 Integration by Parts LIVE
Partition a product's area into rectangles to see the by-parts formula.
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LIATE: a rule of thumb for choosing u
When in doubt, pick u as the function that comes first in LIATE: Logarithmic, Inverse-trig, Algebraic (powers of x), Trigonometric, Exponential. Whatever is earlier in that list differentiates toward something simpler, so let it be u; the rest becomes dv. It works because differentiating a log or a power tends to simplify, while sines and exponentials integrate just as easily as they differentiate.
📝 Worked example: Evaluate ∫ x eˣ dx.
- By LIATE the algebraic factor x comes before the exponential, so let \( u=x \) and \( dv=e^{x}\,dx \).
- Then \( du=dx \) and \( v=e^{x} \).
- Apply the formula: \( \int x e^{x}\,dx = x e^{x}-\int e^{x}\,dx \).
- The new integral is trivial: \( \int e^{x}\,dx=e^{x} \).
✓ ∫ x eˣ dx = <strong>x eˣ − eˣ + C = eˣ(x − 1) + C</strong>.
📝 Worked example: Evaluate ∫ ln x dx.
- There seems to be only one factor, but write it as \( \ln x \cdot 1 \) and take \( u=\ln x,\ dv=dx \).
- Then \( du=\tfrac{1}{x}\,dx \) and \( v=x \).
- By parts: \( \int \ln x\,dx = x\ln x-\int x\cdot\tfrac{1}{x}\,dx = x\ln x-\int 1\,dx \).
✓ ∫ ln x dx = <strong>x ln x − x + C</strong>.
Definite integrals by parts
For a definite integral you evaluate the uv term at both limits and subtract, then integrate what remains. The boundary term carries the limits just like any other evaluated antiderivative.
\[ \int_{a}^{b} u\,dv = \Big[uv\Big]_{a}^{b}-\int_{a}^{b} v\,du \]
The definite version: evaluate the uv boundary term across the limits, then subtract the remaining integral.
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Don't forget v du — and don't loop forever
Two common traps. First, the formula subtracts ∫v du, not ∫u dv again — keep track of which piece is which. Second, a poor choice of u can send you in circles, giving back the original integral or something worse; if that happens, swap your choice of u and dv. (Sometimes a deliberate loop is the method, as with ∫ex sin x dx, where the original integral reappears and you solve for it algebraically.)
Check your understanding
1. The integration-by-parts formula is:
Integrating the product rule (uv)′ = u′v + uv′ and rearranging gives ∫u dv = uv − ∫v du.
2. Integration by parts is the reverse of which differentiation rule?
It comes directly from integrating the product rule, (uv)′ = u′v + uv′.
3. For ∫ x cos x dx, which choice of u does LIATE recommend?
Algebraic (x) comes before Trigonometric (cos x) in LIATE, so let u = x; then du = dx and dv = cos x dx, which simplifies the integral.
4. What is ∫ x eˣ dx?
With u = x, dv = eˣ dx: ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.
5. What is ∫ ln x dx?
Take u = ln x, dv = dx, so v = x and du = dx/x: x ln x − ∫x·(1/x) dx = x ln x − x + C.
✅ Key takeaways
- Integration by parts is the product rule integrated: ∫u dv = uv − ∫v du.
- Split the integrand into u (to differentiate) and dv (to integrate); a good split makes ∫v du simpler.
- LIATE (Log, Inverse-trig, Algebraic, Trig, Exponential) is a reliable guide for choosing u.
- ∫x eˣ dx = eˣ(x − 1) + C and ∫ln x dx = x ln x − x + C are the canonical worked examples.
- For definite integrals, evaluate the [uv] boundary term across the limits, then subtract ∫v du.