Substitution with Linear Inner Powers

Why Linear Inner Expressions Need Substitution

A power such as $(3x-2)^4$ is not a power of $x$: its base changes with $x$. The antiderivative power rule applies most directly when the variable itself is the base. Substitution lets us treat the entire linear expression as a new variable, apply that rule, and then express the answer in terms of $x$.

The crucial detail is that changing the variable also changes the differential. If the inner expression is $ax+b$, its derivative is the constant $a$. The differential therefore contributes a factor of $1/a$ when we solve for $dx$. Losing that factor gives an answer whose derivative is too large or too small.

Definition: A linear inner expression is an expression of the form $ax+b$, where $a$ and $b$ are constants. For substitution with this inner expression, set $u=ax+b$, so $du=a\,dx$ and $dx=\frac{1}{a}\,du$, provided $a\ne 0$.

The Power Formula for a Linear Base

Suppose $a\ne 0$, and consider an integral whose integrand is $(ax+b)^p$. Substituting $u=ax+b$ means that the whole base becomes $u$, while $dx$ becomes $\frac{1}{a}\,du$. Thus the transformed integral is:

\int (ax+b)^p\,dx = \frac{1}{a}\int u^p\,du.

When $p\ne -1$, the power rule for antiderivatives gives $\int u^p\,du=\frac{u^{p+1}}{p+1}+C$. Substituting $ax+b$ back for $u$ produces the formula below.

Theorem: If $a\ne 0$ and $p\ne -1$, then on any interval where the real-valued expression $(ax+b)^p$ is defined, an antiderivative is $$ \int (ax+b)^p\,dx = \frac{(ax+b)^{p+1}}{a(p+1)}+C. $$ For an integrand $c(ax+b)^p$, where $c$ is a constant, multiply the result by $c$. The interval must avoid points where the integrand is undefined. For arbitrary real $p$, the usual real-valued power is defined when $ax+b>0$; integer powers may also be defined for negative bases, provided a negative exponent does not encounter a zero base.

The exponent $p=-1$ is the exception to the power rule: it would require division by $p+1=0$. In that case, use the logarithmic antiderivative instead. On any interval where $ax+b\ne 0$,

\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b|+C.

For a constant numerator $c$, the result is $\frac{c}{a}\ln|ax+b|+C$. The absolute value allows the logarithm to be defined on either side of the point where the linear expression is zero, as long as the interval does not include that point.

Worked Examples

Worked Example: A Positive Integer Power

Find $\int 4(2x+3)^5\,dx$.

Set $u=2x+3$. Then $du=2\,dx$, so $dx=\frac{1}{2}\,du$. Transform both the power and the differential, keeping the factor $4$:

\begin{aligned} \int 4(2x+3)^5\,dx &= \int 4u^5\left(\frac{1}{2}\,du\right)\\ &=2\int u^5\,du\\ &=2\left(\frac{u^6}{6}\right)+C\\ &=\frac{u^6}{3}+C\\ &=\frac{(2x+3)^6}{3}+C. \end{aligned}

Check by differentiating the answer:

\frac{d}{dx}\left(\frac{(2x+3)^6}{3}\right) = \frac{1}{3}\cdot 6(2x+3)^5\cdot 2 = 4(2x+3)^5.

This matches the integrand, so the antiderivative is correct.

Worked Example: A Fractional Power

Find $\int 6(3x+1)^{-1/2}\,dx$. The exponent is fractional, so work on an interval where $3x+1>0$, ensuring the expression is real-valued.

Let $u=3x+1$. Then $du=3\,dx$, so $dx=\frac{1}{3}\,du$. Substitute throughout the integral:

\begin{aligned} \int 6(3x+1)^{-1/2}\,dx &= \int 6u^{-1/2}\left(\frac{1}{3}\,du\right)\\ &=2\int u^{-1/2}\,du\\ &=2\left(\frac{u^{1/2}}{1/2}\right)+C\\ &=4u^{1/2}+C\\ &=4\sqrt{3x+1}+C. \end{aligned}

Differentiate to check:

\frac{d}{dx}\left(4\sqrt{3x+1}\right) = 4\cdot\frac{1}{2}(3x+1)^{-1/2}\cdot 3 = 6(3x+1)^{-1/2}.

The result differentiates to the original integrand on the stated domain.

Worked Example: A Negative Slope and a Negative Power

Find $\int (5-2x)^{-3}\,dx$. The base must not be zero, so consider an interval that does not contain $x=\frac{5}{2}$.

Set $u=5-2x$. Then $du=-2\,dx$, so $dx=-\frac{1}{2}\,du$. The negative sign matters:

\begin{aligned} \int (5-2x)^{-3}\,dx &= -\frac{1}{2}\int u^{-3}\,du\\ &= -\frac{1}{2}\left(\frac{u^{-2}}{-2}\right)+C\\ &= \frac{1}{4}u^{-2}+C\\ &= \frac{1}{4(5-2x)^2}+C. \end{aligned}

For the check, differentiate the final expression, keeping track of the derivative of the inner expression:

\begin{aligned} \frac{d}{dx}\left(\frac{1}{4}(5-2x)^{-2}\right) &= \frac{1}{4}\cdot(-2)(5-2x)^{-3}\cdot(-2)\\ &= (5-2x)^{-3}. \end{aligned}

This is the original integrand. The two negative factors in the derivative cancel, consistent with the positive sign in the antiderivative.

Why the Formula Works

The substitution method here follows from the chain rule. To see why the coefficient $a$ must appear in the denominator, start with the proposed antiderivative for $p\ne -1$ and differentiate it. The chain rule differentiates the outer power and then multiplies by the derivative $a$ of the inner expression:

\begin{aligned} \frac{d}{dx}\left(\frac{(ax+b)^{p+1}}{a(p+1)}\right) &= \frac{1}{a(p+1)}(p+1)(ax+b)^p\cdot a\\ &=(ax+b)^p. \end{aligned}

The factor $a$ from differentiating the inner expression cancels the $a$ in the denominator, and $p+1$ cancels as well. This verifies the formula directly. It also shows why the formula requires $a\ne 0$ and $p\ne -1$: those values would make its denominator zero.

Substitution gives the same reasoning in integral form. The expression $u=ax+b$ changes the base to $u$, and $dx=\frac{1}{a}\,du$ accounts for the constant rate at which that base changes with $x$. Applying the ordinary power rule to $u^p$ and then returning to $x$ is therefore not a separate trick; it is the chain rule run in reverse.

When There Is an Extra Constant Factor

In some integrals, a constant multiplies the power. It must stay in the calculation along with the factor from $dx$. For example, $\int c(ax+b)^p\,dx$ becomes $\frac{c}{a}\int u^p\,du$, not merely $\int u^p\,du$. For $p\ne -1$, this gives $\frac{c(ax+b)^{p+1}}{a(p+1)}+C$. When $c=a$, the constant factors cancel, which is why an integrand with the inner derivative as its coefficient may look especially direct.

The same principle applies when the coefficient does not match the inner derivative. Keep the factor rather than trying to make it disappear. The resulting constant is part of the antiderivative and is necessary for differentiation to recover the original integrand.

A Common Mistake: Forgetting to Change $dx$

Consider $\int (4x+1)^2\,dx$. A common incorrect calculation is to set $u=4x+1$, write $\int u^2\,du$, and conclude that the answer is $\frac{(4x+1)^3}{3}+C$. This treats $dx$ as though it were $du$, even though $du=4\,dx$.

The derivative exposes the error:

\frac{d}{dx}\left(\frac{(4x+1)^3}{3}\right) = \frac{1}{3}\cdot 3(4x+1)^2\cdot 4 = 4(4x+1)^2,

which is four times the intended integrand. Correct the substitution by using $dx=\frac{1}{4}\,du$:

\begin{aligned} \int (4x+1)^2\,dx &=\frac{1}{4}\int u^2\,du\\ &=\frac{u^3}{12}+C\\ &=\frac{(4x+1)^3}{12}+C. \end{aligned}

Now check:

\frac{d}{dx}\left(\frac{(4x+1)^3}{12}\right) = \frac{1}{12}\cdot 3(4x+1)^2\cdot 4 = (4x+1)^2.

The corrected answer differentiates to the original integrand. This check is especially useful when a linear expression has a coefficient other than $1$, or when its slope is negative.

What to Keep Track Of

For an integral involving a power of $ax+b$, the reliable procedure is to name the entire inner expression, solve for $dx$, and transform every factor in the integrand. Then apply the power rule, except when the exponent is $-1$, and substitute the original expression back. On a valid interval, differentiating the result should reproduce the integrand exactly.

Pay particular attention to the domain. A negative integer power is undefined where its base is zero. A fractional power may require a positive base to remain real-valued. An antiderivative formula is valid on intervals where the integrand is defined; do not extend it across a point where the original expression is undefined.

Check Your Understanding

For each question, keep the differential factor in the substitution and check antiderivatives by differentiating.

Find $\int (3x+4)^2\,dx$, and verify your result.
Find $\int 8(2x-1)^{-1/2}\,dx$ on an interval where the integrand is real-valued.
Find $\int (7-3x)^{-2}\,dx$, and check the sign by differentiation.
Why is the power formula for $(ax+b)^p$ not valid when $p=-1$? What form applies instead?
A student substitutes $u=5x+2$ but replaces $dx$ by $du$. Explain the missing factor and how it affects the answer.

Substitution with Linear Inner Powers

What You'll Learn

Why Linear Inner Expressions Need Substitution

The Power Formula for a Linear Base