Substituting the Linear Radicand

Why Substitute the Radicand?

When an integrand contains both \(x\) and a square root such as \(\sqrt{ax+b}\), those two quantities are linked: the radicand is a linear function of \(x\). Using that radicand as a new variable can simplify the relationship. In particular, the substitution \(t=\sqrt{ax+b}\) expresses both \(x\) and \(dx\) using powers of \(t\).

This is useful when the integrand is a rational expression in \(x\) and \(\sqrt{ax+b}\). After substitution, its algebraic factors become polynomials in \(t\), so the transformed integrand is a rational function of \(t\), wherever its denominators are nonzero.

Deriving the Substitution Formulas

Starting from \(t=\sqrt{ax+b}\), square both sides. Since \(t\) is defined as the nonnegative square root, this gives \(t^2=ax+b\). Solving for \(x\) gives the first required formula. Differentiating that formula gives the second.

The restriction \(a\ne 0\) is essential: otherwise \(ax+b\) does not determine \(x\) from \(t\) by this formula. For real \(t\), the original radicand must satisfy \(ax+b\ge 0\). The substitution uses only \(t\ge 0\); allowing negative \(t\) would describe a different branch and would not equal the principal square root.

For example, if the integrand contains a rational expression \(R(x,\sqrt{ax+b})\), then replace \(x\) by \((t^2-b)/a\), replace \(\sqrt{ax+b}\) by \(t\), and multiply by \(dx=(2t/a)\,dt\). If \(R\) is rational in its two arguments, these replacements produce a rational function of \(t\), provided the resulting denominators are not zero at the points under consideration.

Why the Change of Variable Works

The formulas above are not merely convenient algebra. They describe the same integral in a different variable. On any interval where \(ax+b>0\), the function \(t=\sqrt{ax+b}\) is differentiable and has derivative \(dt/dx=a/(2t)\), which is nonzero. Thus it can be reversed there, and the chain rule accounts for the differential factor.

To see directly how antiderivatives are preserved, let \(G(t)\) be an antiderivative of the transformed integrand, and define \(F(x)=G(\sqrt{ax+b})\). The chain rule gives \(F'(x)=G'(t)\,dt/dx\). The transformed integrand includes the factor \(dx/dt\), so the factors \(dt/dx\) and \(dx/dt\) cancel. Consequently, differentiating the result after substituting back gives the original integrand.

Worked Indefinite Integrals

Definite Integrals and Their Bounds

For a definite integral, transform the bounds as well as the integrand. If \(x\) runs from \(x_1\) to \(x_2\), the corresponding \(t\)-bounds are \(\sqrt{ax_1+b}\) and \(\sqrt{ax_2+b}\). When \(a<0\), \(t\) decreases as \(x\) increases, so the new lower bound may be larger than the new upper bound. This is correct: the oriented integral automatically accounts for the reversal.

A Common Mistake: Forgetting the Differential

A frequent error is to replace the root and \(x\), but leave out the factor from \(dx\). In the first worked example, the incorrect calculation might be

This omits \(dx=t\,dt\). It would produce \(t^2/4-\tfrac12\ln t+C\), which is not the answer. Indeed, differentiating that expression after substituting \(t=\sqrt{2x+3}\) gives \(1/2-1/(2t^2)\), not the original integrand \((x+1)/t\). The correction is to replace the entire differential: multiply the transformed expression by \(t\,dt\). That gives \(\tfrac12\int(t^2-1)\,dt\), as computed above.

When This Substitution Is Useful

The method is especially effective when replacing \(x\) and the square root makes the integrand rational in \(t\). The transformed rational integral can then be handled with the rational-function integration methods appropriate to its factors. The substitution does not guarantee that every integral becomes easy, but it provides a systematic way to remove the linear square root from the algebra.

Keep the domain in view. The original square root requires \(ax+b\ge 0\), and any denominator in the integrand must remain nonzero on the interval being integrated. The inverse formula \(x=(t^2-b)/a\) follows from \(t^2=ax+b\), but the restriction \(t\ge0\) is what ensures that \(t\) represents the specified square root. If a definite interval includes a point where \(ax+b=0\), the substitution can be applied first on an interior interval where the radicand is positive; an endpoint case must then be interpreted by taking a limit if the integrand is improper.