Polynomial Definite Integrals By Substitution

From a Composite Integrand to a Simpler Integral

A polynomial composite can look complicated because the same expression appears repeatedly inside a power or another polynomial. If the integrand also contains the derivative of that inner expression, substitution replaces the repeated expression with a new variable. For a definite integral, the bounds must change along with the variable.

The goal is not to expand every power and integrate term by term. Instead, identify a polynomial $q(x)$ whose derivative is present as a factor, set $u=q(x)$, transform the endpoints, and integrate the resulting polynomial in $u$.

Core pattern: If the integrand has the form $P(q(x))q'(x)$, where $P$ and $q$ are polynomials, then the substitution $u=q(x)$ turns the integral into an integral of $P(u)$, with bounds $q(a)$ and $q(b)$.

The Definite-Integral Substitution Formula

For a polynomial $P$, an antiderivative $F$ is a function satisfying $F'(u)=P(u)$. The Fundamental Theorem of Calculus evaluates the integral of $P$ from one endpoint to another as the difference of $F$ at those endpoints.

Theorem: Let $q$ be continuously differentiable on $[a,b]$, and let $P$ be a polynomial. Then $$ \int_a^b P(q(x))q'(x)\,dx = \int_{q(a)}^{q(b)}P(u)\,du. $$ The right-hand integral is oriented: if $q(b)<q(a)$, its upper endpoint is smaller than its lower endpoint, and the usual sign convention for definite integrals applies. No assumption that $q$ is increasing or decreasing is required.

For the polynomial substitutions in this tutorial, $q$ is a polynomial, so it is continuously differentiable. The integrand $P(q(x))q'(x)$ is also a polynomial and therefore continuous on the closed interval $[a,b]$. Thus the proper Riemann integral on the left is defined. The transformed integrand is a polynomial as well, so the integral on the right is defined.

Why the Formula Works

Choose an antiderivative $F$ of $P$, so $F'(u)=P(u)$. The chain rule gives the derivative of the composite function $F(q(x))$. Because $q$ and $F$ are differentiable, and $q$ is continuously differentiable here, this composite is continuously differentiable on $[a,b]$.

\frac{d}{dx}F(q(x)) = F'(q(x))q'(x) = P(q(x))q'(x).

The Fundamental Theorem of Calculus can therefore be applied on $[a,b]$: the integral of this continuous derivative equals the change in $F(q(x))$ between the endpoints. Applying the same theorem to $P$ on the transformed interval gives the claimed identity.

\begin{aligned} \int_a^b P(q(x))q'(x)\,dx &=F(q(b))-F(q(a))\\ &=\int_{q(a)}^{q(b)}P(u)\,du. \end{aligned}

This argument also explains the endpoint rule: the original endpoints $a$ and $b$ become $q(a)$ and $q(b)$. The chain rule supplies the derivative factor, and the Fundamental Theorem accounts for the endpoint values.

A Reliable Procedure

Choose the inner polynomial.
Look for a repeated polynomial expression $q(x)$ inside the integrand.

Match its derivative.
Differentiate $q(x)$ and check that the integrand contains $q'(x)$ as a factor, including any constant multiplier.

Transform both bounds.
Evaluate $q(a)$ and $q(b)$. These become the lower and upper bounds in the new variable, respectively.

Integrate and evaluate.
Write the resulting polynomial in $u$, find an antiderivative, and subtract its value at the lower transformed bound from its value at the upper transformed bound.

The differential notation $du=q'(x)\,dx$ records the derivative factor and the change of variable. For a definite integral, it is essential to pair that change of variable with the transformed bounds; one must not leave the old $x$-bounds attached to an integral in $u$.

Worked Examples

Worked Example: A Power of a Quadratic

Evaluate $\int_0^2 2x(x^2+1)^2\,dx$.

Choose $u=x^2+1$. Then $du=2x\,dx$, which is exactly the remaining factor. Transform the bounds: when $x=0$, $u=0^2+1=1$; when $x=2$, $u=2^2+1=5$. Therefore,

\begin{aligned} \int_0^2 2x(x^2+1)^2\,dx &=\int_1^5 u^2\,du\\ &=\left[\frac{u^3}{3}\right]_1^5\\ &=\frac{5^3}{3}-\frac{1^3}{3}\\ &=\frac{125-1}{3}\\ &= \frac{124}{3}. \end{aligned}

Check by differentiating the corresponding composite antiderivative: $\frac{d}{dx}\left(\frac{(x^2+1)^3}{3}\right)=\frac{1}{3}\cdot 3(x^2+1)^2\cdot 2x=2x(x^2+1)^2$. Its endpoint difference is $\frac{5^3}{3}-\frac{1^3}{3}=\frac{124}{3}$, confirming the result.

Worked Example: A General Polynomial in the Inner Function

Evaluate $\int_0^1 2x\big((x^2+1)^3-2(x^2+1)+4\big)\,dx$.

Set $u=x^2+1$, so $du=2x\,dx$. The bounds become $u=1$ at $x=0$ and $u=2$ at $x=1$. Each occurrence of $x^2+1$ is replaced by $u$:

\begin{aligned} \int_0^1 2x\big((x^2+1)^3-2(x^2+1)+4\big)\,dx &=\int_1^2 (u^3-2u+4)\,du\\ &=\left[\frac{u^4}{4}-u^2+4u\right]_1^2\\ &=\left(\frac{2^4}{4}-2^2+4\cdot2\right) -\left(\frac{1^4}{4}-1^2+4\cdot1\right)\\ &=(4-4+8)-\left(\frac14-1+4\right)\\ &=8-\frac{13}{4}\\ &=\frac{32}{4}-\frac{13}{4}\\ &=\frac{19}{4}. \end{aligned}

To check, define $F(u)=\frac{u^4}{4}-u^2+4u$. Then $F'(u)=u^3-2u+4$. By the chain rule, $\frac{d}{dx}F(x^2+1)=((x^2+1)^3-2(x^2+1)+4)2x$, the original integrand. Also $F(2)-F(1)=\frac{19}{4}$, as calculated.

Worked Example: A Cubic Inner Function

Evaluate $\int_0^1 3x^2(x^3+2)^4\,dx$.

Choose $u=x^3+2$. Its derivative is $du=3x^2\,dx$, matching the factor outside the fourth power. At $x=0$, $u=0^3+2=2$; at $x=1$, $u=1^3+2=3$. Thus,

\begin{aligned} \int_0^1 3x^2(x^3+2)^4\,dx &=\int_2^3 u^4\,du\\ &=\left[\frac{u^5}{5}\right]_2^3\\ &=\frac{3^5}{5}-\frac{2^5}{5}\\ &=\frac{243-32}{5}\\ &=\frac{211}{5}. \end{aligned}

For a check, differentiate $\frac{(x^3+2)^5}{5}$: $\frac{1}{5}\cdot5(x^3+2)^4\cdot3x^2=3x^2(x^3+2)^4$. The endpoint difference is $\frac{3^5-2^5}{5}=\frac{211}{5}$, agreeing with the substitution.

Matching the Derivative Factor

The substitution is effective when the integrand includes the derivative of the chosen inner polynomial. The factor may be written in a different order, but its value must match. For example, with $q(x)=3x^2+1$, the derivative is $q'(x)=6x$, not $3x$. The coefficient matters.

More generally, if a constant multiple of $q'(x)$ appears, it can be accounted for explicitly. For instance, $3x^2\,dx$ is $du$ for $u=x^3+2$. If instead the integrand contains $x^2\,dx$, then $du=3x^2\,dx$, so $x^2\,dx=\frac13du$. That factor of $\frac13$ must remain in the transformed integral. Substitution does not make unmatched constants disappear.

Takeaway: Before integrating, verify the whole change: the polynomial inside the composite, its derivative factor (with the correct coefficient), and both transformed endpoints.

A Common Bounds Error

Suppose an integral is written in $u$ after setting $u=q(x)$, but its limits are still the original $x$-values. This mixes two variables and changes the problem. The transformed integral must use $q(a)$ and $q(b)$, not $a$ and $b$.

For example, in the first worked example the correct transformed expression is $\int_1^5u^2\,du$. Leaving the original bounds in place would give the incorrect calculation $\int_0^2u^2\,du=\frac83$. The correction is to compute $u=0^2+1=1$ and $u=2^2+1=5$ before integrating. The correct value is $\int_1^5u^2\,du=\frac{124}{3}$, not $\frac83$.

The error is not fixed by merely writing $u=x^2+1$ beside the integral. Once the integrand and differential are expressed in $u$, the endpoints must also be expressed in $u$. Keeping $x$-bounds would be appropriate only if the calculation were returned to the original variable before evaluation.

Why This Method Matters

Expanding a polynomial composite can create many terms, even when its structure is simple. Substitution preserves that structure and often reduces the calculation to a basic polynomial antiderivative. It also makes the role of the endpoints transparent: the answer depends on the values of the inner function at the original endpoints.

The theorem applies even if $q$ is not monotone, provided the stated regularity conditions hold. In that case, the transformed endpoints still appear in their original order, and the integral's orientation handles the sign. For this tutorial, the main practical task remains recognizing and matching the polynomial derivative factor.

Check Your Understanding

For each question, identify the inner polynomial and pay particular attention to the transformed bounds and derivative factor.

Evaluate $\int_0^1 4x(2x^2+3)^2\,dx$ by substitution. What are the transformed bounds?
For $u=x^3-1$, what differential factor $du$ must be present for a direct substitution?
Evaluate $\int_0^2 3x^2(x^3+1)^2\,dx$, showing the endpoint transformations.
Why must both bounds change when the integrand and differential are rewritten in terms of $u$?
Does the substitution formula require $q$ to be monotone on the interval? State the regularity condition used in the theorem.

Polynomial Definite Integrals By Substitution

What You'll Learn

From a Composite Integrand to a Simpler Integral

The Definite-Integral Substitution Formula

Why the Formula Works

A Reliable Procedure