Why Use the Root as a Variable?
Some integrands are not rational functions of \(x\), but become rational once the root appearing in them is treated as a variable. When the root is an \(n\)th root of a linear expression, the substitution \(t=(ax+b)^{1/n}\) replaces the root by \(t\) and the linear expression by \(t^n\). Solving for \(x\) then makes every occurrence of \(x\) rational in \(t\) as well.
The substitution is useful only when the algebra and the real domain are handled together. In particular, the differential must include the factor obtained by differentiating \(x\) with respect to \(t\). For an even root, the chosen real root is nonnegative, so the new variable is restricted accordingly.
The Substitution and Its Domain
Set \(t=(ax+b)^{1/n}\), using the real root wherever it exists. Raising both sides to the \(n\)th power and solving for \(x\) gives the inverse relation. Differentiating that relation gives the transformed differential.
For odd \(n\), every real number has a unique real \(n\)th root, so \(t\) may be any real number for which the transformed expressions are defined. For even \(n\), the real \(n\)th root is defined only when \(ax+b\ge 0\), and the conventional real root satisfies \(t\ge 0\). The original integral may impose further restrictions: for example, a denominator cannot vanish. The substitution does not remove such restrictions; it rewrites them in terms of \(t\).
Since \(a\ne 0\), the inverse formula specifies \(x\) uniquely for each allowable \(t\). The map preserves or reverses orientation according to the sign of \(a\): as \(t\) increases through its allowed range, \(x=(t^n-b)/a\) increases if \(a>0\) and decreases if \(a<0\). For a definite integral, transform both endpoints and retain their resulting order. If an endpoint or an interior point is singular, the integral must be interpreted on intervals where the integrand is defined; an improper integral, when needed, is treated by the appropriate limit.
Why the Integrand Becomes Rational
Suppose the integrand is \(R\bigl(x,(ax+b)^{1/n}\bigr)\). After the substitution, the root becomes \(t\), while \(x\) becomes \((t^n-b)/a\). Both replacements are polynomials or rational expressions in \(t\). The differential contributes another polynomial factor, \((n/a)t^{n-1}\). Substituting all three into the integrand therefore produces a rational function of \(t\), wherever the denominators are nonzero.
The formula follows from replacing \(x\) by its inverse expression, replacing the root by \(t\), and using the derivative of the inverse expression for \(dx\). This is the ordinary change-of-variable rule: if \(x=x(t)\), then the differential factor is \(dx=x'(t)\,dt\). The domain restrictions above ensure that the substitution represents the intended real root; singularities must still be excluded or handled as improper limits.
Worked Examples
Worked Example: A Cube Root with an Odd-Root Domain
Evaluate \(\displaystyle \int \frac{x}{\sqrt[3]{2x+1}}\,dx\), on an interval where the integrand is defined.
Let \(t=\sqrt[3]{2x+1}\). Since this is an odd root, \(t\) can be any real number. From \(t^3=2x+1\), solve for \(x\), then differentiate:
The integrand is undefined when \(t=0\), corresponding to \(x=-\frac12\), so work on an interval that does not cross that point. Substitute into the integral:
Integrating term by term gives
Returning to \(x\), one form of the antiderivative is \(\frac{3}{20}(2x+1)^{5/3}-\frac38(2x+1)^{2/3}+C\). To check, differentiate the expression in \(t\) with respect to \(x\). Since \(dt/dx=2/(3t^2)\),
This agrees with the original integrand wherever it is defined.
Worked Example: A Square Root in a Denominator
Evaluate \(\displaystyle \int \frac{dx}{(x+1)\sqrt{3x+1}}\) on its real domain.
The square root requires \(3x+1\ge0\), so \(x\ge-\frac13\). Set \(t=\sqrt{3x+1}\), which gives \(t\ge0\). From \(t^2=3x+1\),
For the original integrand to be defined, \(x+1\ne0\); on this domain \(x+1\ge\frac23\), so it never vanishes. Substitute each factor:
Using the standard antiderivative of \(1/(t^2+c^2)\), the result is \(\sqrt2\arctan(t/\sqrt2)+C\). Thus an antiderivative in \(x\) is
Check by differentiating. With \(t=\sqrt{3x+1}\), \(dt/dx=3/(2t)\), and the derivative of the proposed answer is
because \(x+1=(t^2+2)/3\). The derivative matches the integrand on its real domain.
Worked Example: A Definite Integral with a Fourth Root
Evaluate \(\displaystyle \int_0^{15}\frac{x}{(x+1)^{1/4}}\,dx\).
On this interval \(x+1>0\), so the real fourth root is defined and nonnegative. Set \(t=(x+1)^{1/4}\). Then \(x=t^4-1\) and \(dx=4t^3\,dt\). Transform the endpoints: when \(x=0\), \(t=1\); when \(x=15\), \(t=2\). Therefore
Evaluate the polynomial integral at both endpoints:
For a check, an antiderivative in \(t\) is \(G(t)=4(t^7/7-t^3/3)\). Its derivative is \(4(t^6-t^2)\), exactly the transformed integrand. Also, with \(t=(x+1)^{1/4}\), \(dt/dx=1/(4t^3)\), so \(G'(t)\,dt/dx=t^3-t=x/(x+1)^{1/4}\). The transformed bounds and derivative both agree with the original integral.
A Common Differential Error
A frequent mistake is to differentiate the radicand but then use that derivative as \(dx/dt\), rather than solving for \(x\) first. For instance, with \(t=\sqrt{2x+1}\), the relation is \(x=(t^2-1)/2\). Therefore \(dx/dt=t\), not \(2t\).
The incorrect differential \(dx=2t\,dt\) would instead produce \(2t+C\). Differentiating that proposed answer with respect to \(x\) gives \(2/t\), whereas the original integrand is \(1/t\). The check exposes the extra factor of \(2\). The reliable procedure is to solve for \(x\) as a function of \(t\), then differentiate that expression.
Using the Method Reliably
The substitution is a change of variable, not a guarantee that the resulting integral is easy. Its immediate benefit is structural: a suitable integrand becomes rational in \(t\), after which methods for rational functions can be used. The new rational function may still require algebraic simplification or partial fractions.
Identify the linear expression \(ax+b\), verify \(a\ne0\), and confirm that the other dependence is rational in \(x\) and its \(n\)th root.
Write \(t^n=ax+b\), solve for \(x\), and differentiate to obtain \(dx=(n/a)t^{n-1}dt\).
For even \(n\), require \(ax+b\ge0\) and \(t\ge0\). For either parity, exclude values where the original or transformed integrand is undefined.
Replace \(x\), the root, the differential, and—if the integral is definite—each endpoint. Keep the transformed bounds in their resulting order.
Differentiate an antiderivative with respect to \(x\), or verify the transformed derivative and endpoint values for a definite integral.
Careful domain tracking matters especially near \(ax+b=0\). A transformed expression can look algebraically simpler while still having an endpoint where the substitution variable is zero or the original integrand is singular. The substitution formula is applied on intervals where the expressions are defined; any singular endpoint requires the corresponding limit rather than an ordinary endpoint evaluation.
Check Your Understanding
For each question, keep both the algebraic substitution and the real domain in view.
- For \(t=(5x-2)^{1/3}\), solve for \(x\) and calculate \(dx\) in terms of \(t\) and \(dt\).
- If \(t=\sqrt[4]{2x+3}\), what restrictions apply to \(x\) and \(t\)?
- Under \(t=(x+4)^{1/5}\), rewrite \(\displaystyle \frac{x^2}{(x+4)^{2/5}}\,dx\) as an expression in \(t\) and \(dt\).
- Why must a definite integral’s endpoints be converted when changing from \(x\) to \(t\)?
- For \(t=\sqrt{3x+1}\), identify the error in claiming \(dx=3t\,dt\). What is the correct differential?
