Substitution with Radical Expressions

Why Radicals Can Fit Substitution

A radical can make an integrand look different from the power-rule expressions you already know. But a root is a fractional power, and that often reveals a familiar substitution pattern: an expression appears beneath the radical, and the remaining factor is its derivative or a constant multiple of it.

For instance, in an expression such as $\sqrt{3x^2+2}$, the quantity $3x^2+2$ is the inner expression. Rewriting the square root as $(3x^2+2)^{1/2}$ makes its power visible. If the integrand also contains $6x$, the derivative of the inner expression, substitution turns the integral into a basic power-rule integral.

Key idea: Rewrite the radical as a fractional power, let the expression beneath it be $u$, and check whether the remaining factor supplies $du$, possibly up to a constant.

Roots as Fractional Powers

For a nonnegative real number $a$, the principal $n$th root is written $\sqrt[n]{a}$. In fractional-power notation, $\sqrt[n]{a}=a^{1/n}$. More generally, when the powers are defined, $\sqrt[n]{a^m}=a^{m/n}$. In particular, a square root becomes a power of $1/2$, a cube root a power of $1/3$, and a reciprocal square root a power of $-1/2$.

\sqrt{g(x)}=[g(x)]^{1/2}, \qquad \frac{1}{\sqrt{g(x)}}=[g(x)]^{-1/2}, \qquad \sqrt[3]{g(x)}=[g(x)]^{1/3}.

For an even root, the real-valued expression requires $g(x)\ge 0$. If the radical is in a denominator, then $g(x)>0$, since division by zero is undefined. Odd roots can also be real for negative inputs. Throughout, use fractional powers only on intervals where the corresponding real-valued expressions are defined.

The Substitution Pattern

Suppose an integrand contains a power of an inner function $g(x)$, and its other factor is $g'(x)$ or a constant multiple of $g'(x)$. Set $u=g(x)$. Then $du=g'(x)\,dx$, so the integral becomes a power of $u$ times $du$. For a power $u^p$, where $p\ne -1$, the power rule gives an antiderivative $u^{p+1}/(p+1)$.

Substitution rule for radical powers: If $g(x)>0$ on an interval and $p\ne -1$, then $$ \int g'(x)[g(x)]^p\,dx = \frac{[g(x)]^{p+1}}{p+1}+C. $$ If the integrand has $k g'(x)$ instead of $g'(x)$, factor out $k$ or account for it when replacing $g'(x)\,dx$ by $du$. The special exponent $p=-1$ gives a logarithmic form rather than this power rule.

The condition $p\ne -1$ matters: the displayed denominator would otherwise be zero, and the power rule does not apply. Logarithmic forms are a separate case. In the radical examples below, the exponents are not $-1$.

Why the Method Works

The substitution rule follows directly from the chain rule. This gives a way to justify the result rather than relying only on a pattern.

Start with a power-rule antiderivative.
For $p\ne -1$, a function whose derivative with respect to $u$ is $u^p$ is $H(u)=u^{p+1}/(p+1)$, because $H'(u)=u^p$.

Compose with the inner function.
Replace $u$ by $g(x)$, giving $H(g(x))=[g(x)]^{p+1}/(p+1)$.

Differentiate the composition.
The chain rule gives $\frac{d}{dx}H(g(x))=H'(g(x))g'(x)=[g(x)]^p g'(x)$.

Identify the integrand.
The derivative is exactly $g'(x)[g(x)]^p$, so the proposed expression is an antiderivative wherever the functions involved are defined.

This proof also explains why the derivative factor is essential. Without it, differentiating the proposed expression produces an extra factor $g'(x)$, so the result generally would not match the integrand.

Worked Examples

Worked Example: A Square Root with an Exact Derivative Factor

Evaluate $\int 6x\sqrt{3x^2+2}\,dx$.

Step 1: Rewrite the radical and identify the inner function.

\sqrt{3x^2+2}=(3x^2+2)^{1/2}, \qquad u=3x^2+2.

Step 2: Differentiate the inner function. Since $\frac{du}{dx}=6x$, we have $du=6x\,dx$. This is exactly the remaining factor and differential in the integral.

Step 3: Substitute and integrate.

\int 6x(3x^2+2)^{1/2}\,dx = \int u^{1/2}\,du = \frac{u^{3/2}}{3/2}+C = \frac{2}{3}u^{3/2}+C.

Step 4: Substitute back and check.

F(x)=\frac{2}{3}(3x^2+2)^{3/2}+C,

and differentiating gives

F'(x) = \frac{2}{3}\cdot\frac{3}{2}(3x^2+2)^{1/2}\cdot 6x = 6x\sqrt{3x^2+2}.

This is the original integrand, so the antiderivative is verified.

Worked Example: A Reciprocal Square Root

Evaluate $\int \frac{4x-1}{\sqrt{2x^2-x+5}}\,dx$.

Step 1: Express the denominator as a power. The reciprocal square root is a power of $-1/2$. Let $u=2x^2-x+5$.

\int (4x-1)(2x^2-x+5)^{-1/2}\,dx.

Step 2: Match the differential. Since $\frac{du}{dx}=4x-1$, we have $du=(4x-1)\,dx$. The radicand is positive for every real $x$: its discriminant is $(-1)^2-4(2)(5)=-39<0$, and its leading coefficient is positive.

Step 3: Integrate in $u$.

\int u^{-1/2}\,du = \frac{u^{1/2}}{1/2}+C = 2u^{1/2}+C.

Step 4: Return to $x$ and verify.

F(x)=2\sqrt{2x^2-x+5}+C,

F'(x) = 2\cdot\frac{1}{2\sqrt{2x^2-x+5}}(4x-1) = \frac{4x-1}{\sqrt{2x^2-x+5}}.

The derivative matches the integrand.

Worked Example: Adjusting for a Constant Multiple

Evaluate $\int x^2\sqrt{x^3+4}\,dx$.

Step 1: Choose the inner function. Write the radical as a fractional power and set $u=x^3+4$.

\sqrt{x^3+4}=(x^3+4)^{1/2}, \qquad du=3x^2\,dx.

The integral contains $x^2\,dx$, which is one third of $du$: $x^2\,dx=\frac13du$.

Step 2: Substitute and integrate.

\int x^2(x^3+4)^{1/2}\,dx = \frac13\int u^{1/2}\,du = \frac13\cdot\frac23u^{3/2}+C = \frac29u^{3/2}+C.

Step 3: Substitute back and check.

F(x)=\frac29(x^3+4)^{3/2}+C,

and

F'(x) = \frac29\cdot\frac32(x^3+4)^{1/2}\cdot3x^2 = x^2\sqrt{x^3+4}.

The factors simplify because $\frac29\cdot\frac32\cdot3=1$. Thus the derivative is the original integrand.

Worked Example: A Cube Root

Evaluate $\int (3x^2+2x)\sqrt[3]{x^3+x^2+7}\,dx$ for $x\ge 0$.

Step 1: Convert the root and identify $u$. For $x\ge0$, the radicand is positive. Set $u=x^3+x^2+7$, so $du=(3x^2+2x)\,dx$.

\sqrt[3]{x^3+x^2+7} = (x^3+x^2+7)^{1/3}.

Step 2: Integrate the resulting power.

\int u^{1/3}\,du = \frac{u^{4/3}}{4/3}+C = \frac34u^{4/3}+C.

Step 3: Substitute back and check.

F(x)=\frac34(x^3+x^2+7)^{4/3}+C,

and the chain rule gives

F'(x) = \frac34\cdot\frac43(x^3+x^2+7)^{1/3}(3x^2+2x) = (3x^2+2x)\sqrt[3]{x^3+x^2+7}.

Again, the derivative reproduces the given integrand.

Common Pitfall: Losing the Power or the Derivative Factor

A frequent error is to see a radical and write down a similar-looking radical as its antiderivative, without increasing the power or accounting for the derivative of the inner function. For example, a tempting but incorrect answer to $\int 6x\sqrt{3x^2+2}\,dx$ is $\sqrt{3x^2+2}$. Differentiating that guess reveals the problem:

\frac{d}{dx}\sqrt{3x^2+2} = \frac{6x}{2\sqrt{3x^2+2}} = \frac{3x}{\sqrt{3x^2+2}},

which is not $6x\sqrt{3x^2+2}$. The correction is to integrate the power $u^{1/2}$: its exponent increases to $3/2$, and division by $3/2$ supplies the factor $2/3$. The verified answer is $\frac23(3x^2+2)^{3/2}+C$.

When checking your work, differentiate the complete proposed antiderivative. This simultaneously tests the fractional-power rule, the constant factor, and whether the inner derivative has been included correctly.

When This Pattern Applies

Substitution with a radical is appropriate when the expression under the root can serve as $u$, and the rest of the integrand provides its derivative, perhaps after a constant is factored out. The method does not claim that every radical integral has this pattern. If the derivative factor is absent or does not match, the direct substitution described here may not turn the integral into a simple power of $u$.

Rewrite the radical as a fractional power before integrating.
Differentiate the proposed inner function explicitly.
Match all of its derivative factor, including any constant multiple.
Use the power rule only when its exponent is not $-1$.
Respect the real domain of the root and any denominator.
Differentiate the final answer to confirm it reproduces the integrand.

The central benefit is that a complicated-looking radical expression can become a straightforward power integral once its inner structure is recognized. The check by differentiation confirms that the substitution preserved every factor.

Check Your Understanding

For each question, identify the inner function and show how its derivative relates to the remaining factor.

Rewrite $\sqrt[4]{x^2+1}$ as a fractional power.
Evaluate $\int 8x(x^2+3)^{1/2}\,dx$, then check by differentiation.
Evaluate $\int \frac{3x^2}{\sqrt{x^3+2}}\,dx$. State why the radicand permits the real-valued integrand on its domain.
For $\int (2x+1)(x^2+x+4)^{1/3}\,dx$, what substitution converts the integral to a power of $u$, and what is the resulting antiderivative?
Explain why the formula for integrating $u^p$ used here cannot be applied when $p=-1$.