Integrate by substitution

• Choose an inner expression whose derivative appears as a factor in the integrand.
• Transform every part of an indefinite integral into the substitution variable.
• Back-substitute and verify the antiderivative by differentiation.

Build the idea

Worked example

1. Step 1

   Let u equal three x squared plus one. Its differential is six x d x, which is already the remaining factor in the integral.

   $$u=3x^2+1,\quad du=6x\,dx$$

2. Step 2

   The integral becomes u to the fourth d u. Increase the exponent to five and divide by five.

   $$\int u^4\,du=\frac{u^5}{5}+C$$

3. Step 3

   Replace u with the original quadratic. The constant of integration remains because all antiderivatives differ by a constant.

   $$\frac{(3x^2+1)^5}{5}+C$$

• Replacing the inner expression with u while leaving an unmatched x elsewhere.

  After substitution, every variable factor must be expressed through u and d u; otherwise the transformation is incomplete.

• Forgetting the constant multiplier needed to match d u.

  Compare the inner derivative with the available factor and multiply or divide by constants explicitly before integrating.

Find an antiderivative of two x times the square root of the quantity x squared plus four, end quantity, and verify it.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Lesson overview

   Substitution reverses the chain rule by treating a repeated inner expression as a new variable and matching its derivative elsewhere in the integrand. A chain-rule derivative contains an outer derivative evaluated at an inner expression times the inner derivative. Substitution recognizes that same paired structure inside an integral. Choose an inner expression whose derivative appears as a factor in the integrand.

2. Part 2

   Visual model and equations

   Naming the inner expression u is only useful when d u accounts for the remaining x-dependent factor, possibly after extracting a constant. Imagine x-values passing through an inner function to create u-values. The factor d u over d x rescales small x-widths into u-widths so the outer accumulation can be integrated directly. The integral of f of g of x times g prime of x d x becomes the integral of f of u d u. u equals g of x, and d u equals g prime of x d x.

   • The integral of f of g of x times g prime of x d x becomes the integral of f of u d u.
   • u equals g of x, and d u equals g prime of x d x.
3. Part 3

   Worked example

   Find an antiderivative of six x times the quantity three x squared plus one, end quantity, to the fourth power. Choose the repeated inner quadratic as u, match its derivative six x d x exactly, integrate the resulting power, and back-substitute. Let u equal three x squared plus one. Its differential is six x d x, which is already the remaining factor in the integral. The integral becomes u to the fourth d u. Increase the exponent to five and divide by five. Replace u with the original quadratic. The constant of integration remains because all antiderivatives differ by a constant. An antiderivative is one fifth times the quantity three x squared plus one, end quantity, to the fifth, plus C. Differentiating applies the chain rule: one fifth times five times the fourth power times six x, which simplifies to the original integrand.

   • u equals three x squared plus one, and d u equals six x d x.
   • The integral of u to the fourth d u equals u to the fifth over five plus C.
   • The quantity three x squared plus one, end quantity, to the fifth, divided by five, plus C.
4. Part 4

   Checks, practice, and scope

   A transformed integral must contain only u and d u. After integrating, replace u with its original expression and differentiate the result to check the pattern. Replacing the inner expression with u while leaving an unmatched x elsewhere. After substitution, every variable factor must be expressed through u and d u; otherwise the transformation is incomplete. Forgetting the constant multiplier needed to match d u. Compare the inner derivative with the available factor and multiply or divide by constants explicitly before integrating. Find an antiderivative of two x times the square root of the quantity x squared plus four, end quantity, and verify it. Use u equals x squared plus four, so d u equals two x d x, then integrate u to the one-half power. The antiderivative is two thirds times the quantity x squared plus four, end quantity, raised to the three-halves, plus C. A useful substitution identifies an inner expression and matches its derivative factor. The transformed integral should contain only the new variable and its differential. Differentiating the final answer is the most direct check of an indefinite integral. Not every integral fits a single substitution, and some require algebraic rewriting, integration by parts, partial fractions, or numerical methods. Definite-integral substitution also changes bounds or requires back-substitution before endpoint evaluation.

Scope and limitations

• Not every integral fits a single substitution, and some require algebraic rewriting, integration by parts, partial fractions, or numerical methods.
• Definite-integral substitution also changes bounds or requires back-substitution before endpoint evaluation.