Integration of UV Formula — ∫u dv = uv − ∫v du

A Formula for Products That Refuse to Integrate

Differentiating a product is easy — the product rule handles $x\sin x$ or $xe^x$ in seconds. Integrating one is where students hit a wall, because there is no "product rule for integrals." The integration of uv formula is the closest thing: it doesn't integrate the product directly, it trades a hard integral for an easier one.

The integration of uv formula — also called integration by parts or the uv rule of integration — says:

$$\boxed{;\int u,dv = uv - \int v,du;}$$

You split the integrand into a part $u$ (which you'll differentiate) and a part $dv$ (which you'll integrate), and the formula rewrites the original integral as a product minus a new integral — one you choose your split to make simpler.

What Is the Integration of UV Formula?

The integration of uv formula is the technique for integrating a product of two functions when no straightforward antiderivative exists. It is the integral analogue of the product rule. In its most-used form:

$$\int u,dv = uv - \int v,du.$$

Here $u$ and $v$ are functions of $x$; you pick $u$ and $dv$ from the integrand, then compute $du$ (by differentiating $u$) and $v$ (by integrating $dv$). Written fully in terms of $x$, with $u$ and $w$ the two factors:

$$\int u,w,dx = u\int w,dx - \int\left(u'\int w,dx\right)dx.$$

The point is never to integrate the product head-on. It is to swap the original integral for $uv - \int v,du$, where the new integral $\int v,du$ is simpler than the one you started with. Choose the split so that differentiating $u$ shrinks it.

How Is the Integration of UV Formula Derived?

It comes straight from the product rule for derivatives. Start with two functions $u$ and $v$:

$$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.$$

Integrate both sides with respect to $x$. The left side integrates back to $uv$:

$$uv = \int u,\frac{dv}{dx},dx + \int v,\frac{du}{dx},dx = \int u,dv + \int v,du.$$

Now rearrange to isolate the integral you want:

$$\int u,dv = uv - \int v,du.$$

That's the whole derivation. Integration by parts is the product rule run backwards and solved for one of its two integrals. (The product rule it inverts is the one from our uv differentiation formula guide — worth reading alongside this, because the contrast is the clearest way to see what each rule does.)

How Do You Choose u? The ILATE Rule

The formula only helps if the new integral $\int v,du$ is easier than the original. The choice of $u$ decides that. The ILATE rule (also written LIATE) ranks function types by how good a choice they make for $u$ — pick $u$ from whichever type comes first:

The logic: $u$ should get simpler when differentiated, and $dv$ should stay manageable when integrated. Logarithms and inverse-trig functions simplify dramatically under differentiation (they turn into algebraic fractions), so they make excellent $u$. Exponentials and trig functions barely change when integrated, so they make safe $dv$. For $\int x\ln x,dx$, L beats A, so $u = \ln x$.

Examples of the Integration of UV Formula

Example 1

Evaluate $\int x,e^x,dx$.

By ILATE, Algebraic (A) beats Exponential (E), so $u = x$ and $dv = e^x,dx$. Then $du = dx$ and $v = e^x$:

$$\int x,e^x,dx = uv - \int v,du = x,e^x - \int e^x,dx = x,e^x - e^x + C.$$

Final answer: $\int x,e^x,dx = e^x(x - 1) + C$.

Example 2

Evaluate $\int x\cos x,dx$.

The instinct here is to make the trig function $u$, because trig "feels like the main function." Follow that and watch what happens.

Wrong path. Let $u = \cos x$ and $dv = x,dx$. Then $du = -\sin x,dx$ and $v = \frac{x^2}{2}$:

$$\int x\cos x,dx = \frac{x^2}{2}\cos x - \int \frac{x^2}{2}(-\sin x),dx = \frac{x^2}{2}\cos x + \frac12\int x^2\sin x,dx.$$

The new integral $\int x^2\sin x,dx$ is worse than the original — the power went up from $x$ to $x^2$. The split made the problem harder, not easier.

The break. ILATE puts Algebraic (A) before Trigonometric (T). Choosing the trig function as $u$ violates the rule, and the symptom is exactly this: the algebraic power grows instead of shrinking.

The rescue. Let $u = x$ (Algebraic) and $dv = \cos x,dx$. Then $du = dx$ and $v = \sin x$:

$$\int x\cos x,dx = x\sin x - \int \sin x,dx = x\sin x + \cos x + C.$$

Final answer: $\int x\cos x,dx = x\sin x + \cos x + C$. Differentiating $u = x$ down to $1$ is what collapses the new integral to something trivial.

Example 3

Evaluate $\int x^2\ln x,dx$.

By ILATE, Logarithmic (L) beats Algebraic (A), so $u = \ln x$ and $dv = x^2,dx$. Then $du = \frac{1}{x},dx$ and $v = \frac{x^3}{3}$:

$$\int x^2\ln x,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac{1}{x},dx = \frac{x^3}{3}\ln x - \frac13\int x^2,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C.$$

Final answer: $\int x^2\ln x,dx = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$.

Example 4

Evaluate $\int \ln x,dx$ (a single function, no obvious product).

The trick: write $\ln x$ as a product with $1$. Let $u = \ln x$ and $dv = 1,dx$. Then $du = \frac{1}{x},dx$ and $v = x$:

$$\int \ln x,dx = x\ln x - \int x\cdot\frac{1}{x},dx = x\ln x - \int 1,dx = x\ln x - x + C.$$

Final answer: $\int \ln x,dx = x\ln x - x + C$. Integration by parts cracks "non-product" integrals like $\ln x$ and $\arctan x$ by pairing them with an invisible factor of $1$.

Example 5

Evaluate $\int \arctan x,dx$.

Same $1$-trick: $u = \arctan x$ (Inverse trig, top of ILATE), $dv = 1,dx$, so $du = \frac{1}{1 + x^2},dx$ and $v = x$:

$$\int \arctan x,dx = x\arctan x - \int \frac{x}{1 + x^2},dx.$$

The remaining integral is a standard substitution ($w = 1 + x^2$): $\int \frac{x}{1+x^2},dx = \frac12\ln(1 + x^2)$. So:

$$\int \arctan x,dx = x\arctan x - \frac12\ln(1 + x^2) + C.$$

Final answer: $\int \arctan x,dx = x\arctan x - \dfrac12\ln(1 + x^2) + C$.

Example 6

Evaluate $\int e^x\sin x,dx$ (the loop that closes on itself).

Neither factor simplifies under differentiation, so apply the formula twice and let the original integral reappear. Let $I = \int e^x\sin x,dx$. With $u = \sin x$, $dv = e^x,dx$:

$$I = e^x\sin x - \int e^x\cos x,dx.$$

Apply parts again to $\int e^x\cos x,dx$ with $u = \cos x$, $dv = e^x,dx$:

$$\int e^x\cos x,dx = e^x\cos x + \int e^x\sin x,dx = e^x\cos x + I.$$

Substitute back: $I = e^x\sin x - (e^x\cos x + I) = e^x\sin x - e^x\cos x - I$. Solve for $I$:

$$2I = e^x(\sin x - \cos x) ;\Rightarrow; I = \frac{e^x(\sin x - \cos x)}{2} + C.$$

Final answer: $\int e^x\sin x,dx = \dfrac{e^x(\sin x - \cos x)}{2} + C$. The reappearing integral isn't a dead end — it's solved like an algebraic unknown.

Where the Integration of UV Formula Carries Real Weight

Integration by parts is not just an exam technique — it is one of the load-bearing tools of applied mathematics.

• Physics — quantum mechanics. Deriving expectation values and the momentum operator's action relies on integration by parts to shift a derivative off the wavefunction. The "boundary terms" that drop out are exactly the $uv$ part of the formula.

• Signal processing — the Fourier and Laplace transforms. Computing the transform of a derivative, $\int f'(t)e^{-st},dt$, uses integration by parts to convert differentiation into multiplication by $s$ — the property that makes Laplace transforms solve differential equations.

• Probability and statistics. The expected value $E[X] = \int_0^\infty (1 - F(x)),dx$ for a non-negative random variable is derived by integrating $\int x f(x),dx$ by parts.

• Engineering — the weak form of differential equations. The finite element method, behind nearly all structural and fluid simulation, rewrites differential equations in "weak form" using integration by parts so that solutions need only be once-differentiable.

For Class 12 and first-year calculus students, the immediate use is evaluating integrals like $\int x e^x,dx$ — but the same formula reappears across physics, transforms, and the numerical methods behind modern engineering software.

Tripping Points to Avoid

Mistake 1: Choosing u and dv backwards

Where it slips in: Picking $u$ as the function that grows under differentiation (a trig or exponential factor) instead of the one that shrinks.

Don't do this: Choose $u$ by which function "looks more important." That has no bearing on whether the new integral is easier.

The correct way: Use ILATE — pick $u$ from the type highest on the list.

Mistake 2: Dropping the minus sign before the second integral

Where it slips in: Writing $\int u,dv = uv + \int v,du$ — a plus where the formula needs a minus.

Don't do this: Treat the formula as symmetric. The minus sign is not optional; it comes directly from rearranging the product rule.

The correct way: Always $\int u,dv = uv - \int v,du$. The memorizer archetype, who recites "uv minus integral v du," still loses the sign mid-computation when the line gets long — write the minus first, before filling in the integral.

Mistake 3: Forgetting the constant of integration

Where it slips in: Reporting the final antiderivative without $+ C$, especially after a multi-step problem like Example 6.

Don't do this: Assume the constant doesn't matter because "it's just a constant." On a definite integral it cancels, but on an indefinite one it is part of the answer.

The correct way: Every indefinite integral ends in $+ C$. The silent-understander archetype does the calculus perfectly and then loses a mark on the missing constant, every single time.

Conclusion

• The integration of uv formula is $\int u,dv = uv - \int v,du$ — the integration counterpart of the product rule.

• It is derived by integrating the product rule $\frac{d}{dx}(uv) = u,dv/dx + v,du/dx$ and solving for one integral.

• Use the ILATE rule (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) to choose $u$ so the new integral is simpler.

• The pairing-with-1 trick integrates lone functions like $\ln x$ and $\arctan x$; the loop trick handles $\int e^x\sin x,dx$.

• The most common errors are choosing $u$ and $dv$ backwards, dropping the minus sign, and forgetting $+ C$.

Practice These Before Moving On

1. Evaluate $\int x\sin x,dx$.

2. Evaluate $\int x,\ln x,dx$.

3. Evaluate $\int e^x\cos x,dx$ using the loop method.

If Problem 1 gave a higher power of $x$ in the new integral, return to Mistake 1 — ILATE puts Algebraic before Trigonometric, so $u = x$.

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