Recipe

Integrate by parts

Use when: the integrand is a product of two dissimilar functions (a polynomial times an exponential, logarithm, inverse trig, or trig function) with no substitution that removes it in one step

Goal: the original integral traded for a boundary term plus a strictly easier integral, ready to finish or repeat

Taught in Integration by parts · a later section extends this recipe with more cases

Quick recipe

Rank the factors with LIATE and pick u
Compute $du$ and $v$
Apply the by-parts formula
Evaluate the leftover integral and simplify
Check by differentiating the result

1
List the integrand's two factors and rank them by LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential). Call the higher-ranked factor $u$ , and call the rest of the integrand, together with $dx$ , $dv$ .

Example

For $\int x\cos(3x)\,dx$ : $x$ is Algebraic and $\cos(3x)$ is Trigonometric. Algebraic outranks Trigonometric on LIATE, so $u=x$ and $dv=\cos(3x)\,dx$ .

Why: choosing $u$ by LIATE picks the factor that gets simpler under differentiation, which is what makes the traded integral easier instead of harder.
2
Differentiate $u$ to get $du$ . Integrate $dv$ to get $v$ , dropping the constant of integration here since it gets absorbed later.

Example

$u=x \Rightarrow du=dx$ . $dv=\cos(3x)\,dx \Rightarrow v=\int\cos(3x)\,dx=\dfrac{\sin(3x)}{3}$ .

Why: these two pieces, $du$ and $v$ , are exactly the ingredients the by-parts formula needs, and nothing else is computed at this stage.
3
$\int u\,dv = uv-\int v\,du$ . Write down $uv$ first, then subtract the new integral.

Example

$\int x\cos(3x)\,dx = x\cdot\frac{\sin(3x)}{3} - \int\frac{\sin(3x)}{3}\,dx$
- the new integral $\int v\,du$ is visibly simpler than the original — continue to Step 4 and finish it directly.
- the new integral is exactly as hard, or harder, than the original — the split was backwards; swap which factor is $u$ and restart from Step 1.
Why: this line is just the product rule, integrated and rearranged, so it always holds; whether it helps depends entirely on the split chosen in Step 1.
4
Finish $\int v\,du$ with whatever technique it needs, then combine it with the boundary term from Step 3.

Example

$\int\frac{\sin(3x)}{3}\,dx = -\frac{\cos(3x)}{9}+C_1,\qquad \int x\cos(3x)\,dx = \frac{x\sin(3x)}{3}+\frac{\cos(3x)}{9}+C$
- Not yet: the leftover integral is again a product of the same two kinds of factors, needing another by-parts split (like $x^3e^x$ , which needs three rounds) — covered in Repeated parts and tabular integration
Why: the leftover integral almost never comes for free; it still has to be solved before the original problem is finished.
5
Differentiate the final answer and confirm it reproduces the original integrand exactly.

Example

$\dfrac{d}{dx}\left[\dfrac{x\sin(3x)}{3}\right]=\dfrac{\sin(3x)}{3}+x\cos(3x)$ and $\dfrac{d}{dx}\left[\dfrac{\cos(3x)}{9}\right]=-\dfrac{\sin(3x)}{3}$ . Adding them cancels the $\sin(3x)/3$ terms and leaves $x\cos(3x)$ , matching the original integrand.

Why: a sign error in Step 3's subtraction or an arithmetic slip in Step 4 is otherwise invisible until this check catches it.

Common mistakes

Choosing $u$ to be whichever factor looks easier to integrate, instead of whichever simplifies more under differentiation (LIATE ranks by the latter).
Forgetting the boundary term $uv$ and writing the new integral alone.
Dropping the minus sign in $uv-\int v\,du$ .
Letting $u$ equal the entire integrand with $dv=dx$ , which almost never simplifies anything.
Swapping $u$ and $dv$ after a bad first attempt without recomputing $du$ and $v$ to match.