Integration by Substitution: Match the Inner Function and Its Derivative

Substitution reverses the chain rule. If u = g(x), replace g′(x)dx by du as well as replacing the inner expression. Rewrite the entire integral in one variable before integrating.

Look for a complete derivative pair

An inner expression and its derivative often appear together in an integrand. A denominator such as 3x² + 4 has derivative 6x, so a numerator 6x suggests a logarithm after substitution. This is more informative than choosing the longest expression simply because it looks complicated.

Constants matter. If the numerator is x instead of 6x, then x dx = du/6. Keep that factor outside the new integral. Replacing only the denominator while leaving x dx unchanged mixes variables and produces an incomplete substitution.

Worked example

Find ∫6x/(3x² + 4) dx. Let u = 3x² + 4, so du = 6x dx. The integral becomes ∫1/u du, giving ln|u| + C. Since 3x² + 4 is positive for every real x, the result can be written ln(3x² + 4) + C.

Differentiate the answer to check it: the chain rule gives (1/(3x² + 4)) × 6x, exactly the original integrand. This check verifies both the chosen inner expression and the numerical factor.

Definite limits must follow the new variable

For ∫ from 0 to 1 of 2x/(x² + 1) dx, choose u = x² + 1. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫ from 1 to 2 of 1/u du = ln 2.

Alternatively, find ln(x² + 1) first and then use the original x-limits. Both approaches are valid, but do not insert x-limits into an antiderivative written in u. A final numerical definite integral does not need an added arbitrary constant.

When the pattern does not fit immediately

For ∫x(x² + 1)³ dx, the derivative of the inner expression is 2x, so a factor 1/2 is needed. After substitution, the integral is (1/2)∫u³ du = u⁴/8 + C. Differentiation confirms the denominator 8, which is easy to miss if you ignore the constant.

Common mistake

Do not try to apply substitution by replacing symbols without accounting for dx. Write the differential relation on a separate line and ensure no old-variable factor remains unless you have expressed it in terms of u.

Also retain interval restrictions for logarithmic antiderivatives. The formula ∫1/u du = ln|u| + C works on intervals excluding zero; it does not allow a definite integral to cross a singularity without further analysis.

u ke saath du bhi badlo; aadha substitution karke mat rukna.

Three practice questions with worked answers

1. Evaluate ∫2x(x² + 1)³ dx.

   Set u = x² + 1 and du = 2x dx. Then ∫u³ du = u⁴/4 + C, giving (x² + 1)⁴/4 + C.

2. Evaluate ∫cos x/(2 + sin x) dx.

   Let u = 2 + sin x, so du = cos x dx. The answer is ln(2 + sin x) + C because the argument stays positive. Angles use radians.

3. Evaluate ∫ from 0 to 1 of 2x/(x² + 1) dx.

   The substitution u = x² + 1 changes the limits to 1 and 2. Thus the value is [ln u]₁² = ln 2.

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Substitution. No textbook or exam-board affiliation is claimed.