Products Of Powers With A Shared Inner Function

Why Shared Powers Simplify an Integral

Products of powers can make an integrand look more complicated than it is. If every power has the same inner expression, the product can be combined into a single power. When the derivative of that inner expression is also present, substitution reduces the integral to a familiar power-rule or logarithmic form.

For instance, the factors $(x^2+1)^2$ and $(x^2+1)^{-1/3}$ have the same base. Their product is a single power of $x^2+1$, because multiplying powers with the same positive base means adding their exponents. The additional factor $2x$ is the derivative of $x^2+1$, which makes the substitution direct.

Central idea: Combine powers of a shared inner function by adding their exponents. Then use the derivative of that inner function as the differential in a substitution.

Combining Powers with a Shared Base

For a positive real number $b$ and real exponents $a_1,\ldots,a_n$, the product law for powers is

b^{a_1}b^{a_2}\cdots b^{a_n} = b^{a_1+a_2+\cdots+a_n}.

Thus, if the same positive expression $g(x)$ appears in several factors, the exponents add. Define their total by $r=a_1+\cdots+a_n$. Then the product is $g(x)^r$, provided the powers are real-defined and the product law applies. Requiring $g(x)>0$ throughout the interval is a convenient condition that guarantees this for arbitrary real exponents.

Definition: For a product of powers with a shared inner function $g(x)$, the total exponent is the sum $r=a_1+\cdots+a_n$ of the exponents on that shared base.

After combining the factors, look for $g'(x)$, possibly multiplied by a constant, elsewhere in the integrand. If the integrand contains exactly $g'(x)$ times a function of $g(x)$, the substitution $u=g(x)$, $du=g'(x)\,dx$, converts the integral to one in $u$.

The General Antiderivative Rule

Suppose $g$ is differentiable on an interval and $g(x)>0$ there. Let $a_1,\ldots,a_n$ be real numbers and let $r=a_1+\cdots+a_n$. Combining the powers and using $u=g(x)$ gives the following rule.

Theorem: On an interval where $g$ is differentiable and positive, $$ \int g'(x)\prod_{j=1}^{n}g(x)^{a_j}\,dx = \begin{cases} \dfrac{g(x)^{r+1}}{r+1}+C, & r\ne -1,\\[6pt] \ln(g(x))+C, & r=-1, \end{cases} \qquad r=\sum_{j=1}^{n}a_j. $$

The formula assumes the derivative factor is $g'(x)$ itself. If the integrand instead has $c\,g'(x)$, where $c$ is a constant, that constant remains as a multiplier in the result. The rule follows from the ordinary power rule when $r\ne-1$, and from the derivative of the natural logarithm when $r=-1$.

Why the Rule Works

The justification is a direct use of the chain rule. First, on a positive interval, the product law combines the factors:

\prod_{j=1}^{n}g(x)^{a_j} = g(x)^{a_1+\cdots+a_n} = g(x)^r.

For $r\ne-1$, differentiate the proposed antiderivative. The chain rule gives

\frac{d}{dx}\left(\frac{g(x)^{r+1}}{r+1}\right) = \frac{1}{r+1}(r+1)g(x)^r g'(x) = g'(x)g(x)^r.

This is the original integrand after the powers have been combined, so the proposed expression is an antiderivative.

When $r=-1$, positivity permits the logarithm rule:

\frac{d}{dx}\ln(g(x)) = \frac{g'(x)}{g(x)} = g'(x)g(x)^{-1}.

That is again the combined integrand. The two cases differ because the power-rule denominator $r+1$ would be zero when $r=-1$; the logarithm gives the appropriate antiderivative instead.

Worked Examples

Worked Example: A Noninteger Total Exponent

Evaluate

\int (x^2+1)^2(x^2+1)^{-1/3}\,2x\,dx.

Step 1: Combine the powers. The shared inner function is $g(x)=x^2+1$, which is positive for every real $x$. The total exponent is

r=2+\left(-\frac13\right) =\frac63-\frac13 =\frac53.

So the integrand becomes $2x(x^2+1)^{5/3}$. Since $g'(x)=2x$, the derivative factor is present.

Step 2: Apply the rule. Here $r+1=\frac53+1=\frac83$, so

\int 2x(x^2+1)^{5/3}\,dx = \frac{(x^2+1)^{8/3}}{8/3}+C = \frac38(x^2+1)^{8/3}+C.

Check: Differentiate the result:

\frac{d}{dx}\left(\frac38(x^2+1)^{8/3}\right) = \frac38\cdot\frac83(x^2+1)^{5/3}\cdot 2x = 2x(x^2+1)^{5/3}.

Since $(x^2+1)^{5/3}=(x^2+1)^2(x^2+1)^{-1/3}$, this derivative equals the original integrand.

Worked Example: The Logarithmic Case

Evaluate the integral on any interval that does not contain $x=\frac23$:

\int (3x-2)^4(3x-2)^{-5}\,3\,dx.

Step 1: Combine the powers. Both exponents are integers, so the factors are real-defined wherever $3x-2\ne0$. Their total exponent is

$$ r=4+(-5)=-1. $$

The product of the powers is $(3x-2)^{-1}$. Also, the derivative of $g(x)=3x-2$ is $g'(x)=3$. Thus the integral is

\int \frac{3}{3x-2}\,dx.

Step 2: Use the logarithmic form. On either interval separated by $x=\frac23$, the result is

\ln|3x-2|+C.

The absolute value is appropriate because $3x-2$ can be negative, while the combined integrand $3/(3x-2)$ is real-defined on either interval that avoids its zero.

Check: Differentiating gives

\frac{d}{dx}\ln|3x-2| = \frac{3}{3x-2} = 3(3x-2)^{-1} = 3(3x-2)^4(3x-2)^{-5}.

This agrees with the original integrand wherever $3x-2\ne0$.

Worked Example: A Definite Integral

Evaluate

\int_0^1 2x(x^2+3)^{3/4}(x^2+3)^{-1/4}\,dx.

Step 1: Combine the powers. Since $x^2+3>0$, the real powers are defined throughout $[0,1]$. Their total exponent is

r=\frac34-\frac14=\frac24=\frac12.

The integral becomes $\int_0^1 2x(x^2+3)^{1/2}\,dx$, and $2x$ is the derivative of $x^2+3$.

Step 2: Find an antiderivative. Since $r+1=\frac32$,

\int 2x(x^2+3)^{1/2}\,dx = \frac{(x^2+3)^{3/2}}{3/2} = \frac23(x^2+3)^{3/2}.

Step 3: Evaluate at the endpoints.

\begin{aligned} \int_0^1 2x(x^2+3)^{1/2}\,dx &= \left[\frac23(x^2+3)^{3/2}\right]_0^1\\ &= \frac23\left((1^2+3)^{3/2}-(0^2+3)^{3/2}\right)\\ &= \frac23\left(4^{3/2}-3^{3/2}\right)\\ &= \frac23(8-3\sqrt3)\\ &= \frac{16}{3}-2\sqrt3. \end{aligned}

Check: The derivative of $\frac23(x^2+3)^{3/2}$ is $\frac23\cdot\frac32(x^2+3)^{1/2}\cdot2x=2x(x^2+3)^{1/2}$, which is the combined integrand. The original factors multiply to that same power because $\frac34-\frac14=\frac12$. The endpoint subtraction therefore evaluates the stated definite integral.

A Common Exponent Error

When integrating a power, it is not enough to combine the exponents in the product. After finding the total exponent $r$, the power antiderivative uses $r+1$ as its exponent and divides by $r+1$. Forgetting this second step produces a different derivative.

Common mistake: For the first example, one might combine the powers correctly but then write $\frac{(x^2+1)^{5/3}}{5/3}$ as the antiderivative. This is wrong: its derivative is $2x(x^2+1)^{2/3}$, not $2x(x^2+1)^{5/3}$. The correction is to raise the exponent by one: $$ \int 2x(x^2+1)^{5/3}\,dx = \frac{(x^2+1)^{8/3}}{8/3}+C. $$

A second condition matters in the logarithmic case. For arbitrary real exponents, one cannot assume that individual powers of a negative base are real-defined or that they may be combined using the same product law. The logarithmic expression $\ln|g(x)|$ is a valid antiderivative of $g'(x)/g(x)$ wherever $g(x)\ne0$. But when starting from a product of powers, first ensure the original factors are real-defined and their exponent law applies. The integer-power example above satisfies that condition away from the zero of its base.

Applying the Rule to Definite Integrals

For a definite integral, the same combination and antiderivative steps apply. If $g$ is continuously differentiable and positive on $[a,b]$, the integrand is continuous there, and the Fundamental Theorem of Calculus allows the antiderivative to be evaluated at the endpoints. For $r\ne-1$, this gives

\int_a^b g'(x)\prod_{j=1}^{n}g(x)^{a_j}\,dx = \frac{g(b)^{r+1}-g(a)^{r+1}}{r+1}.

For $r=-1$ under the same positivity condition, the value is $\ln(g(b))-\ln(g(a))$. If the combined integrand is instead $g'(x)/g(x)$, and $g$ is nonzero on the whole interval, an antiderivative is $\ln|g(x)|$, giving $\ln|g(b)|-\ln|g(a)|$. The nonzero condition matters: an interval containing a zero of the denominator does not satisfy this formula as an ordinary integral without further analysis.

A Reliable Calculation Pattern

For products of powers with a shared inner function, organize the calculation in this order:

Name the shared inner function.
Write it as $g(x)$, and identify where the powers are real-defined.

Add the exponents.
Compute $r=a_1+\cdots+a_n$ and replace the product by $g(x)^r$.

Match the derivative factor.
Check whether the remaining factor is $g'(x)$, or a constant multiple of it.

Use the correct antiderivative case.
For $r\ne-1$, use $g^{r+1}/(r+1)$; for $r=-1$, use the logarithmic form under its domain conditions.

Verify.
Differentiate the proposed antiderivative and compare it with the original product, or evaluate the antiderivative at the stated bounds.

This pattern matters because it separates two different jobs: simplifying the product of powers and integrating the resulting composite power. Keeping the jobs distinct makes the exponent arithmetic easier to check and helps reveal when the logarithmic case applies.

Check Your Understanding

For each question, identify the shared inner function and total exponent before applying an integration rule.

What is the total exponent in $(x+4)^{5/2}(x+4)^{-3/2}$?
For $\int 4x(x^2+1)^{-2}(x^2+1)^3\,dx$, what is the combined integrand and which antiderivative case applies?
Why does the antiderivative rule for a total exponent $r\ne-1$ use $r+1$, rather than $r$, as the new exponent?
On what domain is $\int 5/(5x-1)\,dx=\ln|5x-1|+C$ valid?
What condition ensures that arbitrary real powers of $g(x)$ in a product are real-defined throughout an interval?