Why Polynomial Substitution Works
Some integrands contain a polynomial raised to a power, together with a factor that resembles the derivative of that polynomial. For example, the factor \(x\) in \(x(x^2+3)^4\) is closely related to the derivative of \(x^2+3\), which is \(2x\). That relationship makes substitution effective: the polynomial becomes a single variable, and its derivative factor becomes the differential.
The important step is not merely to spot a polynomial power. It is to check whether the rest of the integrand supplies the derivative of the inside expression, possibly after adjusting by a constant. Once the factors match, the power rule for antiderivatives can be applied to the new variable.
The Substitution Pattern
Let \(P\) be a differentiable polynomial, and suppose that the powers in the expression are defined on the interval being considered. Set \(u=P(x)\). Then \(du=P'(x)\,dx\). When the integrand contains \(P'(x)\,dx\), the substitution changes the remaining power of \(P(x)\) into a power of \(u\).
For noninteger real powers, the expression must be defined on the interval in question; requiring \(P(x)>0\) throughout the interval is a sufficient condition. The formula excludes \(n=-1\), because the power rule would require division by zero. That case has a logarithmic form, rather than the power formula given here.
In practice, the derivative factor may not be written exactly as \(P'(x)\). A constant multiple is enough. For instance, if the inside expression is \(ax^2+b\), its derivative is \(2ax\). Therefore, an integrand with a factor of \(x\) can match the derivative after multiplying or dividing by the constant \(2a\).
This quadratic formula is valid wherever the power is defined. It follows because \(d(ax^2+b)=2ax\,dx\), so \(x\,dx\) is one-half of \(1/a\) times that differential.
Worked Examples
Worked Example: A Quadratic Inside with a Matching Factor
Evaluate \(\int 6x(3x^2+2)^5\,dx\).
The inner polynomial is \(3x^2+2\). Its derivative is \(6x\), exactly the factor outside the power. Let
Substitute both expressions into the integral:
Apply the power rule for antiderivatives:
Check by differentiating the result. The chain rule gives
This is the original integrand, so the antiderivative is correct.
Worked Example: Adjusting for a Missing Constant
Evaluate \(\int x(2x^2-1)^3\,dx\).
The inside polynomial is \(2x^2-1\), whose derivative is \(4x\). The integrand has \(x\,dx\), which is one-fourth of \(du\). Set
Substitution gives
Replace \(u\) with the original polynomial:
Check by differentiation:
The factors \(4\) from the power rule and \(4x\) from differentiating the inside cancel the denominator \(16\), leaving the required factor \(x\).
Worked Example: A General Polynomial Inside
Evaluate \(\int (3x^2-4x)(x^3-2x^2+7)^4\,dx\).
Here the inner expression is cubic rather than quadratic. Differentiate it:
The derivative is exactly the other factor in the integrand. Set
Then
Substituting back gives
Check by differentiating:
This matches the original integrand exactly.
Why the Formula Is True
The substitution rule is not a separate trick disconnected from differentiation. It reverses the chain rule. The chain rule says that differentiating a power of a composite function produces two factors: the power rule contributes a factor involving the inner function, and differentiating the inner function contributes its derivative.
For \(n\ne -1\), consider \(F(x)=\frac{[P(x)]^{n+1}}{n+1}\), on an interval where the power is defined.
The derivative of the outer power contributes \((n+1)[P(x)]^n\), and the derivative of the inside contributes \(P'(x)\).
The factor \(n+1\) cancels the denominator, leaving \(F'(x)=P'(x)[P(x)]^n\), the original integrand.
This verifies the theorem directly. If a constant \(k\) multiplies the integrand, multiplying the proposed antiderivative by \(k\) makes its derivative \(kP'(x)[P(x)]^n\), as required. The constant of integration \(C\) accounts for the fact that any two antiderivatives on an interval differ by a constant.
How to Recognize the Match
When the inside expression is a polynomial, differentiate it before trying to integrate. Then compare its derivative with the factors outside the power. The comparison should include the coefficient, not just the variable terms.
- If the derivative appears exactly, use it as \(du\).
- If the derivative appears up to a constant factor, adjust by that constant.
- If the derivative factor is absent or does not match, this substitution pattern does not directly reduce the integral to a power of \(u\).
For example, the derivative of \(x^2+5\) is \(2x\). Thus \(x\,dx\) is half of \(d(x^2+5)\). The derivative of \(x^3+5\), however, is \(3x^2\), so a lone factor of \(x\) does not match it. The inner expression and the remaining factors must be considered together.
A Common Mistake: Losing the Constant Factor
A frequent error is to substitute the inside polynomial correctly but forget to convert the differential factor. Consider \(\int x(x^2+1)^3\,dx\). Since \(u=x^2+1\), differentiating gives \(du=2x\,dx\). The incorrect calculation below treats \(x\,dx\) as though it were \(du\):
There is a missing factor of one-half. From \(du=2x\,dx\), it follows that \(x\,dx=\frac12du\), so the corrected calculation is
Differentiate the incorrect answer to see the error:
That result is twice the original integrand. By contrast, differentiating the corrected answer gives
Keeping the differential and its coefficient visible is the simplest way to avoid this mistake.
Why This Pattern Matters
Polynomial substitution lets a complicated-looking expression be treated as a familiar power, provided the derivative of the inside is present. It also makes the relationship between differentiation and integration concrete: the factor \(P'(x)\) is precisely what the chain rule creates when differentiating a power of \(P(x)\).
For AP Calculus problems, the main decision is often whether the factors match. Do not assume that every polynomial power can be handled this way. Check the derivative of the inside, account for constant multipliers, and then confirm the final expression by differentiation. If those checks succeed, the substitution has done more than simplify the notation: it has identified the structure that makes the antiderivative possible.
Check Your Understanding
For each question, identify the inner polynomial and explain how its derivative relates to the remaining factor.
- Evaluate \(\int 4x(x^2+6)^2\,dx\), and check the result by differentiation.
- Evaluate \(\int x(5x^2+3)^4\,dx\). What constant factor is needed when replacing \(x\,dx\)?
- For \(\int (2x+1)(x^2+x+4)^6\,dx\), what should \(u\) be, and what is \(du\)?
- Explain why the power formula in this tutorial does not apply when \(n=-1\).
- A student claims \(\int x(3x^2+1)^2\,dx=\frac{(3x^2+1)^3}{3}+C\). Differentiate the claim and identify the coefficient error.
