Why This Substitution Works
A useful pattern in integration occurs when an expression contains a function raised to a power, multiplied by the derivative of that function. The expression may look complicated because the inner function is itself a polynomial or another familiar function. But if the matching derivative factor is present, the integral can often be reduced to a power of a single variable.
The key is to treat the inner function as a new variable. This is not a trick that changes the integral arbitrarily: the chain rule explains exactly why the resulting antiderivative works. The method applies both to positive integer powers, where the inner function may take any real value, and to general real powers on domains where those powers are defined as real numbers.
The Substitution Formula and Its Conditions
Let \(I\) be an interval and let \(g\) be differentiable on \(I\). If \(n\) is a positive integer, then \(g(x)^n\) is defined for every real value of \(g(x)\). The power rule and chain rule give
The same formula holds for a real exponent \(p\) with \(p\ne -1\) on any interval where \(g(x)>0\). On such an interval, \(g(x)^p\) is defined as a real-valued function, and the result is
The restriction \(p\ne -1\) matters because the power rule would require division by \(p+1\). When \(p=-1\), the integrand is \(g'(x)/g(x)\), which is the logarithmic derivative form and has antiderivative \(\ln|g(x)|+C\) on intervals where \(g\) is nonzero.
For a general real exponent, the condition \(g(x)>0\) is a convenient domain condition that ensures every real power used in the formula is real-valued. Some particular exponents can also be defined for negative inputs, but those cases require separate attention to the exponent and domain. The positive-input condition avoids that ambiguity.
Deriving the Rule from the Chain Rule
The substitution rule follows directly from differentiating the proposed antiderivative. This gives both a justification for the method and a reliable way to verify its result.
For \(p\ne -1\), an antiderivative of \(u^p\) is \(u^{p+1}/(p+1)\), provided the powers are defined on the domain under consideration.
Set \(u=g(x)\), giving the candidate \(F(x)=g(x)^{p+1}/(p+1)\).
By the chain rule, \(F'(x)=g(x)^p g'(x)\). Thus \(F\) is an antiderivative of the integrand on any interval where the formula is valid.
More explicitly, when \(g(x)>0\) and \(p\) is real with \(p\ne-1\),
For a positive integer \(n\), the same calculation is valid even when \(g(x)\) is zero or negative, because the integer powers involved are defined for all real inputs. The chain rule calculation therefore proves the positive-integer version without requiring \(g(x)>0\).
Worked Examples
Worked Example: A Positive Integer Power
Evaluate \(\int 6x(3x^2+2)^4\,dx\). The inner function is \(g(x)=3x^2+2\), and its derivative is \(g'(x)=6x\), exactly the factor multiplying the fourth power.
Set \(u=3x^2+2\). Differentiating gives
Substitute both \(u\) and \(du\) into the integral:
Apply the power rule:
Replace \(u\) by \(3x^2+2\):
Check by differentiating:
The derivative matches the original integrand, so the antiderivative is correct.
Worked Example: A General Real Power
Evaluate \(\int (4x+1)^{3/2}\,dx\) as a real-valued integral. The exponent is a non-integer real number, so use the positive-input domain \(4x+1>0\), which means \(x>-1/4\).
Set \(u=4x+1\). Then \(du=4\,dx\), or \(dx=du/4\). Substitution gives
The exponent \(3/2\) is not \(-1\), so the power rule applies:
Returning to \(x\),
Check by differentiating on that domain:
This agrees with the integrand. The factor \(1/4\) introduced by \(dx=du/4\) is essential.
Worked Example: A Definite Integral with a Negative Exponent
Evaluate \(\int_0^1 2x(1+x^2)^{-1/3}\,dx\). Since \(1+x^2\ge1>0\) throughout the interval, the real power is defined there.
Let \(u=1+x^2\), so \(du=2x\,dx\). Change the bounds as well: when \(x=0\), \(u=1\), and when \(x=1\), \(u=2\). Therefore,
The exponent is \(-1/3\), not \(-1\), so the power rule applies:
To check the antiderivative after substituting back, differentiate:
This is the original integrand, confirming the antiderivative used in the evaluation.
Tracking the Inner Derivative
The substitution \(u=g(x)\) works only when the differential factor is accounted for. If the integrand contains a constant multiple of \(g'(x)\), that constant must remain in the calculation. For example, if the integrand is \(c\,g(x)^p g'(x)\), where \(c\) is constant, then
under the same domain conditions. Equivalently, the constant can be carried through the \(u\)-integral. If the integrand instead contains a factor that is a constant multiple of \(g'(x)\), identify that multiple before integrating; do not silently discard it.
A useful recognition test is to differentiate the proposed inner function. If the integrand contains \(g(x)^p\), calculate \(g'(x)\) and compare it with the remaining factor. Exact agreement makes the substitution direct. A constant mismatch can be adjusted by a constant factor. If the remaining factor is not \(g'(x)\) or a constant multiple of it, this particular power substitution does not directly produce the displayed power integral.
A Common Mistake: Losing a Constant Factor
Consider \(\int (3x-1)^4\,dx\). A common incorrect step is to integrate as if \(u=3x-1\) had derivative \(1\):
Differentiate the proposed answer to see the problem:
The result is three times the required integrand. The correction is to include the factor from the inner derivative. Set \(u=3x-1\), so \(du=3\,dx\) and \(dx=du/3\). Then
Checking the corrected result gives
The derivative now matches. This check is especially valuable when the inner function has a non-unit derivative.
Using the Rule Reliably
For an integral that appears to contain a composite power, proceed in a consistent order:
- Identify the proposed inner function \(g(x)\).
- Differentiate it to find \(g'(x)\).
- Compare the derivative with the factor outside the power, including any constant multiplier.
- Check the exponent and the domain. The power rule requires \(p\ne-1\); general real powers are safely handled where \(g(x)>0\).
- Substitute \(u=g(x)\), integrate the resulting power, and substitute back if the integral is indefinite.
- Differentiate the proposed antiderivative to verify the integrand.
For a definite integral, the same substitution can be used with transformed bounds, or an antiderivative can be substituted into the original bounds. In either approach, the inner function and its derivative must be handled together. The essential fact is the chain rule: the derivative of a power of \(g(x)\) produces both the corresponding power of \(g(x)\) and the factor \(g'(x)\).
Check Your Understanding
For each question, identify the inner function and account for its derivative before applying the power rule.
- Evaluate \(\int 8x(4x^2+1)^3\,dx\), then differentiate your answer to check it.
- For which values of \(x\) is \((2x-5)^{1/2}\) real-valued? On that domain, evaluate \(\int (2x-5)^{1/2}\,dx\).
- Evaluate \(\int (5x+2)^6\,dx\). What factor is required because the inner derivative is \(5\)?
- Why does the power rule in this substitution fail when the exponent is \(-1\), and what standard antiderivative form applies instead when \(g(x)\ne0\)?
- For \(\int_0^1 3x^2(1+x^3)^{1/2}\,dx\), what are the transformed bounds after setting \(u=1+x^3\)?
