Integration by Parts: A Complete Guide to Solving Integrals Using the By‑Parts Method

Introduction

Integration by parts is a powerful technique used to evaluate integrals involving the product of two functions. It is derived from the product rule of differentiation and is especially useful when one part of the integrand becomes simpler after differentiation.

1. The Formula of Integration by Parts

The method is based on the product rule:

d(uv) = u dv + v du

Integrating both sides gives:

∫ u dv = uv − ∫ v du

This is the fundamental formula of integration by parts.

2. How to Choose u and dv

Choosing u and dv correctly is the key to success. A common guideline is the LIATE rule:

• Logarithmic functions → ln(x)
• Inverse trigonometric → arctan(x), arcsin(x)
• Algebraic → x, x²
• Trigonometric → sin(x), cos(x)
• Exponential → eˣ

Choose u as the function that becomes simpler when differentiated.

3. Basic Examples

Example 1: ∫ x eˣ dx

u = x        → du = dx
dv = eˣ dx   → v = eˣ

∫ x eˣ dx = x eˣ − ∫ eˣ dx
           = x eˣ − eˣ + C
           = eˣ (x − 1) + C

Example 2: ∫ x cos(x) dx

u = x        → du = dx
dv = cos(x) dx → v = sin(x)

∫ x cos(x) dx = x sin(x) − ∫ sin(x) dx
               = x sin(x) + cos(x) + C

4. Integration by Parts Twice

Some integrals require applying the method more than once.

Example: ∫ eˣ cos(x) dx

Apply integration by parts twice and solve the resulting equation:

Let I = ∫ eˣ cos(x) dx

u = cos(x), dv = eˣ dx
→ du = −sin(x) dx, v = eˣ

I = eˣ cos(x) + ∫ eˣ sin(x) dx

Apply again:

u = sin(x), dv = eˣ dx
→ du = cos(x) dx, v = eˣ

I = eˣ cos(x) + eˣ sin(x) − ∫ eˣ cos(x) dx

I = eˣ (sin(x) + cos(x)) − I

2I = eˣ (sin(x) + cos(x))

I = (1/2) eˣ (sin(x) + cos(x)) + C

5. Integration by Parts for Logarithmic Functions

Logarithmic functions do not integrate easily, so they are always chosen as u.

Example: ∫ ln(x) dx

u = ln(x)      → du = 1/x dx
dv = dx        → v = x

∫ ln(x) dx = x ln(x) − ∫ x * (1/x) dx
            = x ln(x) − ∫ 1 dx
            = x ln(x) − x + C

6. Integration by Parts for Inverse Trigonometric Functions

Example: ∫ arctan(x) dx

u = arctan(x)     → du = 1/(1 + x²) dx
dv = dx           → v = x

∫ arctan(x) dx = x arctan(x) − ∫ x/(1 + x²) dx
               = x arctan(x) − (1/2) ln(1 + x²) + C

7. Reduction Formulas (Advanced)

Integration by parts can generate recursive formulas for integrals of powers.

Example: ∫ xⁿ eˣ dx

∫ xⁿ eˣ dx = xⁿ eˣ − n ∫ xⁿ⁻¹ eˣ dx

This reduces the power step by step.

8. Integration by Parts in Definite Integrals

∫ab u dv = [uv]ab − ∫ab v du

Example

∫01 x eˣ dx
= [x eˣ]01 − ∫01 eˣ dx
= (1·e − 0) − (e − 1)
= 1

9. When to Use Integration by Parts

• When the integrand is a product of two functions
• When one part becomes simpler after differentiation
• When logarithmic or inverse trig functions appear
• When reduction formulas are needed

Conclusion

Integration by parts is an essential technique for solving integrals involving products, logarithmic functions, inverse trigonometric functions, and exponential expressions. With practice, recognizing when and how to apply this method becomes intuitive and greatly simplifies complex integrals.