Integration by Parts — Formula, Rules & Examples

Integration by Parts

A formula used to integrate the product of two functions.

Formula:

\displaystyle \int u\,dv = uv - \int v\,du

Example 1:

Evaluate

\displaystyle \int xe^{x/2}\,dx\text{.}

Use u = x and dv = e^x/2dx. Then we get du = dx and v = 2e^x/2. This can be summarized:

u = x	dv = e^x/2dx
du = dx	v = 2e^x/2

It follows that

$\begin{aligned} \int xe^{x/2}\,dx &= 2xe^{x/2} - \int 2e^{x/2}\,dx \\ &= 2xe^{x/2} - 4e^{x/2} + C \end{aligned}$

Example 2:

Evaluate

\displaystyle \int \tan^{-1} x\,dx\text{.}

Use the following:

u = tan^-1 x	dv = dx
$\displaystyle du = \frac{1}{1+x^2}dx$	v = x

Thus

$\begin{aligned} \int \tan^{-1} x\,dx &= x\tan^{-1} x - \int \frac{x}{1+x^2}\,dx \\ &= x\tan^{-1} x - \frac{1}{2}\int \frac{2x}{1+x^2}\,dx \\ &= x\tan^{-1} x - \frac{1}{2}\ln\left|1+x^2\right| + C \\ &= x\tan^{-1} x - \frac{1}{2}\ln\left(1+x^2\right) + C \end{aligned}$

Example 3:

Evaluate

\displaystyle \int e^x \sin x \, dx\text{.}

Let I = $\displaystyle \int e^x \sin x \, dx\text{.}$ Proceed as follows:

u = sin x	dv = e^x dx
du = cos x dx	v = e^x

Thus

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

Now use integration by parts on the remaining integral . Use the following assignments:

u = cos x	dv = e^x dx
du = –sin x dx	v = e^x

Thus

$\begin{aligned} \int e^x \sin x \, dx &= e^x \sin x - \left[e^x \cos x - \int e^x (-\sin x) \, dx\right] \\ &= e^x \sin x - e^x \cos x - \int e^x \sin x \, dx \end{aligned}$

Note that $\displaystyle \int e^x \sin x \, dx$ appears on both sides of this equation. Replace it with I and then solve.

$\begin{aligned} I &= e^x \sin x - e^x \cos x - I \\ 2I &= e^x \sin x - e^x \cos x \\ I &= \frac{1}{2}e^x \sin x - \frac{1}{2}e^x \cos x \end{aligned}$

We finally obtain

$\int e^x \sin x \, dx = \frac{1}{2}e^x \sin x - \frac{1}{2}e^x \cos x + C$

See also

Integration methods

Key Formula

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

Where:

$u$ = A function of x that you choose to differentiate
$dv$ = The remaining part of the integrand, which you choose to integrate
$du$ = The derivative of u (times dx)
$v$ = The antiderivative of dv

Worked Example

Problem: Evaluate ∫ x·eˣ dx.

Step 1: Choose u and dv. Let u = x because its derivative simplifies, and let dv = eˣ dx because it is easy to integrate.

u = x, \quad dv = e^x\, dx

Step 2: Differentiate u and integrate dv to find du and v.

du = dx, \quad v = e^x

Step 3: Substitute into the integration by parts formula: ∫ u dv = uv − ∫ v du.

\int x\, e^x\, dx = x\, e^x - \int e^x\, dx

Step 4: Evaluate the remaining integral, which is now straightforward.

= x\, e^x - e^x + C

Step 5: Factor the result for a clean final form.

= e^x(x - 1) + C

Answer:

\int x\, e^x\, dx = e^x(x - 1) + C

Another Example

This example shows how to use integration by parts when the integrand appears to be a single function rather than an obvious product. By setting dv = dx, you create the 'second function' needed for the formula.

Problem: Evaluate ∫ ln(x) dx.

Step 1: This integral has only one obvious function, ln(x). The trick is to set dv = dx so that you can still apply integration by parts.

u = \ln x, \quad dv = dx

Step 2: Differentiate u and integrate dv.

du = \frac{1}{x}\, dx, \quad v = x

Step 3: Apply the formula: ∫ u dv = uv − ∫ v du.

\int \ln x\, dx = x \ln x - \int x \cdot \frac{1}{x}\, dx

Step 4: Simplify and evaluate the remaining integral.

= x \ln x - \int 1\, dx = x \ln x - x + C

Answer:

\int \ln x\, dx = x \ln x - x + C

Frequently Asked Questions

How do you choose u and dv in integration by parts?

A common guideline is the LIATE rule: prioritize choosing u in this order — Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The idea is that u should be the function that becomes simpler when differentiated, while dv should be something you can readily integrate. For example, in ∫ x·eˣ dx, you pick u = x (algebraic) because its derivative is just 1, and dv = eˣ dx because eˣ integrates easily.

When do you need to apply integration by parts more than once?

You apply it multiple times when the first application produces a new integral that is still a product of two functions. A classic case is ∫ x²·eˣ dx, where the first round reduces x² to 2x, and a second round reduces 2x to 2. Another special case is ∫ eˣ·sin x dx, where applying the formula twice brings back the original integral, letting you solve for it algebraically.

What is the difference between integration by parts and u-substitution?

U-substitution reverses the chain rule and works best when the integrand contains a function and its derivative (a composite structure). Integration by parts reverses the product rule and is designed for integrands that are products of two unrelated functions. If the integrand looks like f(g(x))·g'(x), try u-substitution first. If it looks like f(x)·g(x) with no clear chain-rule pattern, integration by parts is usually the right approach.

Integration by Parts vs. U-Substitution

	Integration by Parts	U-Substitution
Reverses which rule	Product rule	Chain rule
Formula	∫ u dv = uv − ∫ v du	∫ f(g(x))·g'(x) dx = ∫ f(u) du
Best used when	Integrand is a product of two different types of functions	Integrand contains a composite function and its inner derivative
Typical examples	∫ x·eˣ dx, ∫ x·sin x dx, ∫ ln x dx	∫ 2x·eˣ² dx, ∫ cos(3x) dx
Complexity	May need to be applied multiple times	Usually applied once

Why It Matters

Integration by parts is one of the most frequently used techniques in any calculus course and appears on nearly every AP Calculus BC and university calculus exam. It is essential for integrating logarithmic, inverse trigonometric, and many product-type functions that no other elementary method can handle. Beyond coursework, it underlies results in physics (such as deriving equations of motion), probability (computing expected values), and engineering (solving differential equations).

Common Mistakes

Mistake: Choosing u and dv in the wrong order, making the integral harder instead of simpler.

Correction: Use the LIATE mnemonic (Logarithmic → Inverse trig → Algebraic → Trigonometric → Exponential) to pick u. The function earlier in this list should generally be u because it simplifies when differentiated. If your new integral is more complex than the original, switch your choices.

Mistake: Forgetting the negative sign in the formula, writing ∫ u dv = uv + ∫ v du instead of uv − ∫ v du.

Correction: The formula comes from the product rule d(uv) = u dv + v du, rearranged to u dv = d(uv) − v du. Integrating both sides gives the minus sign. Always double-check that you have subtracted the second integral.

Related Terms

Integration — The general process that this technique supports
Integration Methods — Overview of all integration techniques
Product Rule — The differentiation rule that integration by parts reverses
Formula — Integration by parts is a core calculus formula
Function — u and v are functions chosen from the integrand
Product — The technique targets products of two functions
Integration by Substitution — Alternative technique reversing the chain rule
Definite Integral — Parts formula extends to definite integrals with boundary terms