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Integration by Parts

In this lesson you’ll learn integration by parts, the technique for integrating products of functions. It’s the product rule in reverse.

When you see an integral like x times e to the x, or x times sin(x), or ln(x), u-substitution won’t help because there’s no chain rule pattern. Instead, you use integration by parts.

The formula comes from reversing the product rule:

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

You split the integrand into two pieces: u (which you’ll differentiate) and dv (which you’ll integrate). The goal is to make the new integral on the right simpler than the original.

Use the LIATE priority for choosing u (pick the first one that appears):

  • Logarithmic (ln x, log x)
  • Inverse trig (arctan, arcsin)
  • Algebraic (polynomials like x, x², x³)
  • Trigonometric (sin, cos, tan)
  • Exponential (e to the x, 2 to the x)

Whatever you pick for u, everything else becomes dv. The idea: u should get simpler when you differentiate it, and dv should be something you can integrate.

If you see a product of two different types of functions (polynomial times exponential, polynomial times trig, logarithm times anything), integration by parts is probably the move. If the integral has a chain rule pattern, try u-substitution first.

Tabular method (repeated integration by parts)

Section titled “Tabular method (repeated integration by parts)”

When u is a polynomial and dv is an exponential or trig function, you’ll need to apply integration by parts multiple times. The tabular method speeds this up:

  1. List successive derivatives of u in one column (until you hit 0)
  2. List successive integrals of dv in another column
  3. Multiply diagonally with alternating signs: +, -, +, -, …
  4. Add them all up

This avoids writing out the formula over and over for integrals like x³ times e to the x.

Example 1: x times e to the x

Evaluate:

xexdx\int x\,e^x\,dx

Pick u = x (algebraic, gets simpler) and dv = e to the x dx (easy to integrate).

u=xdv=exdxu = x \qquad dv = e^x\,dx du=dxv=exdu = dx \qquad v = e^x

Apply the formula:

xexdx=xexexdx=xexex+C\int x\,e^x\,dx = x\,e^x - \int e^x\,dx = x\,e^x - e^x + C

Factor:

=ex(x1)+C= e^x(x - 1) + C

Example 2: x² times e to the x (repeated)

Evaluate:

x2exdx\int x^2\,e^x\,dx

First pass: u = x², dv = e to the x dx.

u=x2dv=exdxu = x^2 \qquad dv = e^x\,dx du=2xdxv=exdu = 2x\,dx \qquad v = e^x =x2ex2xexdx= x^2\,e^x - \int 2x\,e^x\,dx

The remaining integral still needs integration by parts (we just did this one in Example 1):

2xexdx=2(xexex)=2ex(x1)\int 2x\,e^x\,dx = 2(x\,e^x - e^x) = 2e^x(x - 1)

Putting it together:

x2exdx=x2ex2ex(x1)+C=ex(x22x+2)+C\int x^2\,e^x\,dx = x^2\,e^x - 2e^x(x - 1) + C = e^x(x^2 - 2x + 2) + C

Example 3: x times sin(x)

Evaluate:

xsinxdx\int x\,\sin x\,dx

Pick u = x, dv = sin(x) dx.

u=xdv=sinxdxu = x \qquad dv = \sin x\,dx du=dxv=cosxdu = dx \qquad v = -\cos x =xcosx(cosx)dx=xcosx+cosxdx= -x\cos x - \int (-\cos x)\,dx = -x\cos x + \int \cos x\,dx =xcosx+sinx+C= -x\cos x + \sin x + C

Integration by parts shows up whenever you need to integrate a product of different function types:

  • In physics, calculating the work done by a force that depends on position and time simultaneously
  • In probability, finding expected values where you integrate x times a probability density function
  • In signal processing, computing Fourier coefficients requires integrating products of functions with sines and cosines
  • In game development, smooth animation curves often involve polynomial-exponential blends that need this technique

Any time you see “x times something” in an integral, integration by parts is likely your tool.

In the integration by parts formula, what does u represent?
Using LIATE, which function should be u in the integral of x times e to the x?
The integral of x times cos(x) dx equals:
If applying integration by parts makes the integral more complicated, you should:
How many times must you apply integration by parts to evaluate the integral of x² times e to the x dx?