MathIsimple
Back to Calculus
Integration Methods
Core Method

Integration by Parts

Master the standard method for integrating product-type functions

What You'll Learn
  • • Derivation and understanding of the integration by parts formula
  • • LIATE rule: How to choose u and dv
  • • Reduction formulas and cyclic integral techniques
  • • Typical examples and common applications

1. The Formula

Theorem 1.1: Integration by Parts
∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du

This is the reverse of the product rule for differentiation.

LIATE Rule

Choose uu in this priority order:

  • Logarithmic functions
  • Inverse trig functions
  • Algebraic functions (polynomials)
  • Trigonometric functions
  • Exponential functions

2. Examples

Example 2.1: Polynomial times Exponential

Find: ∫xex dx\int xe^x\,dx

Solution:

Let u=xu = x, dv=ex dxdv = e^x\,dx

Then du=dxdu = dx, v=exv = e^x

∫xex dx=xex−∫ex dx=xex−ex+C=ex(x−1)+C\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C
Example 2.2: Logarithmic Function

Find: ∫ln⁡x dx\int \ln x\,dx

Solution:

Let u=ln⁡xu = \ln x, dv=dxdv = dx

Then du=1xdxdu = \frac{1}{x}dx, v=xv = x

∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−x+C\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C
Example 2.3: Repeated Integration by Parts

Find: ∫x2ex dx\int x^2e^x\,dx

Solution:

First application: Let u=x2u = x^2, dv=ex dxdv = e^x\,dx

∫x2ex dx=x2ex−∫2xex dx\int x^2e^x\,dx = x^2e^x - \int 2xe^x\,dx

Second application: For ∫2xex dx\int 2xe^x\,dx, let u=2xu = 2x, dv=ex dxdv = e^x\,dx

∫2xex dx=2xex−2ex+C\int 2xe^x\,dx = 2xe^x - 2e^x + C

Combine:

∫x2ex dx=x2ex−2xex+2ex+C=ex(x2−2x+2)+C\int x^2e^x\,dx = x^2e^x - 2xe^x + 2e^x + C = e^x(x^2-2x+2) + C
Example 2.4: Reduction Formula

Find: ∫exsin⁡x dx\int e^x\sin x\,dx

Solution:

Apply integration by parts twice:

I=∫exsin⁡x dx=exsin⁡x−∫excos⁡x dxI = \int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx
=exsin⁡x−(excos⁡x+∫exsin⁡x dx)= e^x\sin x - (e^x\cos x + \int e^x\sin x\,dx)

Notice ∫exsin⁡x dx\int e^x\sin x\,dx appears again!

I=exsin⁡x−excos⁡x−II = e^x\sin x - e^x\cos x - I
2I=ex(sin⁡x−cos⁡x)2I = e^x(\sin x - \cos x)
I=ex(sin⁡x−cos⁡x)2+CI = \frac{e^x(\sin x - \cos x)}{2} + C

Practice Problems